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Puzzle Corner

Allan J. Gottlieb

Dropping a Rock in

Borneo

Allan J.Gottlieb. 67. is at

the Couranl Institute ol

Mathematical Sciences,

New York University. He

studied mathematics at

M.I.T. and Brandeis.

Send problems,

solutions, and comments

to him at the Courant

Institute, New York

University, 251 Mercer

St, New York, N.Y.,

10012.

Not very much to report this month. I am

very busy helping raise our son David

and managing a research group in "paral

lel processing" (using many cooperating

computers to solve large problems much

faster than is possible with a single

machine). I find both parenthood and

computers to be extremely interesting

and rewarding; I just wish every day were

a little longer so I could spend more time

on each.

Problems

F/M 1 Our first problem for this month is

from Bob Kimble and Jacques Labelle,

who ask that White mate in three:

F/M 2 Next we have a geometry problem

from Greg Huber: Let T be the torus ob

tained by rotating the circle (x - 2) + y =,

1 in the xy-plane about the y-axis. Let P

be a plane tangent to the torus T at the

point (0, 0, 1). Find the volume and sur

face area of the small region obtained by

slicing T with P.

F/M 3 Norman Wickstrand wants you to

solve the simultaneous equations:

X6y = (y2 + 1)x3

ysx =9(xJ + 1)y3

F/M 4 John Prussing adds a new wrinkle

to an old problem:

A pilot flies south over a spherical earth a

distance D, flies due east a distance D,

and then flies due north a distance D, ar

riving back at precisely the starting point.

For D equal to the radius of the earth, find

all solutions for the starting latitude.

F/M 5 Bruce Calder hung a vertical

plumb line down the center of a one-

mile-deep mine shaft somewhere in Bor

neo. He dropped a rock from right next to

the top of the plumb line. How far from

the bottom of the plumb line will the rock

land if we ignore friction?

Speed Department

F/M SD 1 Smith Turner sailed a 10-by-

30-loot boat into a canal lock and then

threw overboard a 10-cubic-foot chest

having a specific gravity of 4. What hap

pened to the water level in the lock? ■

F/M SD 2 A bridge quickie from Doug

Van Patter:

North:

A K7

A 6

V K J 8

4 K 5

* A KQJ7

. South:

A 105 4

V A Q10 4

* 10 7

* 8 6 3

Your partner (North) opened one club

and raised your one-heart bid to four. The

opening lead is 4Q. What play gives you

the best chance of making your contract?

Solutions

OCT1 South is on lead with hearts as trump and is

to take all tricks against the best defense:

♦ K2

V -

♦ J 98

♦ 97

A

5

K Q

10 8

A 43

65

4

1032

The following solution is from Matt BenDaniel. There

are three considerations:

D We must pull East's trumps; this can only be done

by finessing from the dummy.

D The bad offensive fit combined with East's

diamond singleton limit transportation for the offense.

O West must discard the protection from his poten

tial winners.

The basic plan, then, is to get to board in spades, fi

nesse East once, and play another round of hearts.

This will pull East's trumps and will also cause West

to discard. West's discards will determine the de

clarer's play for the rest of the hand. The following

diagram gives the details of play:

A22 FEBRUARY/MARCH 1983

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notes that more "in-ofaer" solutions are possite it

one permits "negative solutions." e.g. -63 = 19 -

82), and Oavid DeWan, who writes, "As a newcomer

to this problem, I was astounded by the variety ol

numbers that can be created with only tour digits, five

operators, and lots of time. I am once again in awe ol

the power of permutation. And genes are permuta

tions of |ust tour "digits" also: A, G, T, and C; no

wonder there are lady bugs, redwoods, penguins,

and porcupines."

A/S 1 Place white bishops on at and a3 and place

black bishops on el and e3. By moving each color

alternately, such that a black bishop can never toko a

white one and conversely, and restricting moves to

ranks 1 to 4 and files a to e (I.e., to ten black

squares), exchange the position of the bishops so

that the black bishops end up on ai and a3 and the

white bishops on el and e3.

Michael Jung and Matthew Fountain seem to have

convincing evidence that no solution is possibte. Howabout it, Mr. Chen?

A/S 2 An isosceles triangle has a bisector ol one of

the two equal angles that is 6 inches long. If the base

is 5 inches, without using trigonometry, find the

length of the two equal sides.

The following solution is from Pnelps Meaker:

The diagram represents the base section of a slen

der isosceles triangle AOD. The base is 5 units, thediagonals AC and BD 6 units; the diagonals bisect

the base angles. Fold the triangle ABD about its side

BD, so that AD lies along CD, extending 5 units to F,

and side AB forms the member BF. Return triangle

ABD to its original position, noting that triangles CBF

and AOD are simBar. Members AB, BC, BF, and CD

are equal. Triangles BCD, BEC, and AED are isos

celes and similar, providing three ratios as follows:

BC/BD = BC/6 (1)

BE/BC = (BD - EDy8C = (6 - AEVBC (2)

AE/AD = AE/5 (3)

Combining (1) and (3):

BC/6 =AE/5;AE =(5 • BCJ/6

Combining (1) and (2):

BC/6 = (6 - AEyBC; iC'/6BC = (36 - 6AEV6BCCombining the above:

B5' = 36 - 6(5 ■ BQ/6; BC* + 5 • BC - 36 = 0;BC = (-5 ± V25 + 4 • 36J/2 = (-5 + 13)/2 = 4.

CD = BC. Since FD = 5 and CD = 4, FC = 1.

In triangle CBF, the sides are four times as long as

the base. Hence the side AO in the similar triangleAOD is tour times AD; and the triangle given in the

problem has sides of 5 4 = 20.

Also solved by Harry Zaremba, Mary Lindenberg.

John Prussing, Everett Leroy, Steven Schmelling.

David Evans, Matthew Fountain, Leon Bankorf, and

the proposer, Emmet Duffy.

A/S 3 Write the prime numbers and take successive

absolute differences:

2 3 5 7 11 13 17 19 23 ...12 2 4 2 4 2 4 ...

Note that the second row starts with a 1. Next repeat

the process of taking successive absolute differ-

A18 JANUARY 198J

ertces, obtaining

2 3 5 7 11 13 17 19 23 29 ...122424246...

1 0 2 2 2 £ 2 2 ...

1 2 0 0 0 0 0 ...

12 0 0 0 0..

1 2 0 0 0 ...

The problem is to prove or disprove that the series ol

leading 1's continues forever.

Matthew Fountain and Winslow Hartford submitted

indications that the sequence of ones continues in

definitely. However, a proof Is still wanting. Note that

the difference between consecutive primes is notbounded (because the density of primes approacheszero).

A/S 4 The "rug puzzle": you want to put wall-to-wall

carpeting into a room that is 9 x 12 feet. You have

two pieces of carpet, one 10 x 10 and the other 1 x

8. These do add to the correct square footage, but

obviously the 10 x 10 must be cut. The challenge is

to devise one continuous cut through the 10 x 10

piece such that the two resulting pieces will exactly fit

the 9 x 12 area with one gap left over into which the

1 x S remnant can fit to complete the job.

The following solution is from Ken Haruta:

Also solved by Fuhsi Ling, Sidney Williams, David

Evans, Matthew Fountain, Avi Omstein, and the proposer, John Fogarty.

A/S 5 One rainy day Fat Timothy was riding a don

key in the countryside ten mJes away from the

nearest shelter when he was caught in a strange rain

which had a uniform mass density and fell straight

down. Worried about being exposed to this strange

precipitation, Timothy rode the faithful donkey as fast

as he could to the nearest shelter. Assuming that thespeed of the donkey was uniform and sufficiently

high, find:

1. The 'amount of rain which fell on Timothy as a

function of 0T. the angle Timothy's body makes withthe ground.

2. The minimum and maximum amount of rain which

might fall on Timothy.

3. The ideal situation in which Timothy would be wet

least.

Approximate Timothy's body as a box.

The following solution is from Harry Zaremba:

In the drawing, assume the dimensions of Timothy's

body are a, b, and c where c is perpendicular to the

page and a < c < b. Also assume that the donkey's

velocity V, is such that the rain falls on the top and

front of Timothy. If V, is the rain's rate of fall, then the

relative velocity of the rain with respect to Timothy is

V = (Vr» + V,»)t, and

4> = tan- (V^V,).

The area of rainfall that will be intercepted byTimothy's body is

A = c(AB + BC) = c(acos (fl - *) + bsin (9 - 4)|,

and the time required to reach the shelter is

T = D/V,,

where D = 10 miles, the distance to the shelter. In

time T, the amount of rain that will fall on the rider is

R =AVT= c(acos (9 - <J) + bsin (9 - <*)]

W,1 + V,» • D/V,.

Noting that

cos * = V,/W,f + V,',

the total amount of rain on the rider becomes

R = (acos (fl - 4,) + bsin (0 - <WJ ■ Dc/eos *. (1)

The minimum amount of rain that can fall on Timothy

occurs when his topside surface is perpendicular to

the apparent path of the rain. In this event, e = <t>, and

the amount of rain from equation (1) becomes

Rmm = Dca/COS *.

For maximum amount of rain, the riding position must

be such that the diagonal OD of Timothy's body is

perpendicular to the relative velocity V of the rain.

Under this condition, the projected area of Timothy's

body on a plane perpendicular to the rain's apparentpath is

A = cVa^FrP.

Thus, the maximum amount of rain is

R-». = Va' + b1 Dc/cos <t>.

The ideal situation for getting wet the least is to rideat an angle where 0 -4> and where the smallest area

of the body is exposed to the rain. In the current

case, the smallest area equals ac, the topside area of

Timothy. When equation (1) is expanded, divided by

cos <t>, and simplified, the result is

R = Dc[ecos 0 + bsin 0 + (asin e - bcos 0) tan *).

It is noted that the term with factor tan <t> will be elimi

nated when the factor (asin 9 - bcos 9) = 0, or tan 6

= b/a. The interesting things about this is that, when

the rider's body makes an angle of 8 = tan-'b/a with

the ground, the amount of rain to fall on the rider re

mains constant, irrespective of the velocity V, of the

donkey. This holds true as long as V, 4 0. The

amount of rain that will fall on the rider is R =Dc(acos 0 + bsin 6).

Also solved by Matthew Fountain, M. Pope, and

Minn Chung, editor of the Physics Department news

letter from which the problem was originally taken.

Better Late Than Never

JAN 2 Henry Frsher has responded.

JUL 3 H. SpacJ has sent photos showing that clocks

in Asia are displayed at 10:09.

Proposers' Solutions to Speed Problems

SD 1 5. Each term Is the number of lit elements

when a calculator displays the digits 0,1,2,3,4,5,6,

7, 8. 9. 0. 1, 2, 3, 4, ...

SD2 Do not ruff the second spade but discard a club

from dummy. Now West cannot force declarer. Take

out two rounds of trumps, then knock out the ♦A.

The VJ lakes the third trump and provides an entry

to dummy's clubs.

James

Goldstein

& Partners

S James Gotdsem 46225 Mibum AvenueMiilbum.NI 0704!(201)4678840

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TECHNOLOGY REVIEW A19

Massachusetts Institute of Technology

Report of the PresidentFor the Academic Year 1981-82

The Institute's principal mission is centered on the

generation of knowledge, and on its preservation and

transmission to new generations of students. During the past

year, there has been broad concern about a number of issues

which remind us, once again, of the power of knowledge,

and of the fact that knowledge is power. The questions of

how closely knowledge is held, how freely it is shared, and

with whom, have permeated most of the major events and

issues df this academic year — linking together several do

mains which, at First glance, may not appear to have much in

common.

These issues include: first, changing patterns in the

organization and support of research; second, questions

relating to the international transfer of technology; and

third, access to higher education. They illustrate the ways in

which the generation, dissemination, and control of

knowledge influence the intellectual and organizational

development of MIT.

The Generation of Knowledge: Changing Patterns of

Research Organization and Support

During the past several years, there has been a steady

increase in the scale of industrial support of sponsored

research in universities. While, on a relative basis, the scale

of such support is still small, it is growing rapidly. At MIT,

for example, industrial support of such research has grown

from $6.7 million in the 1977-78 academic year to $19.7

million in this past year, and now constitutes about a tenth of

the sponsored research conducted on the MIT campus.

For some, the growth in industrial research sponsor

ship raises almost as many questions as opportunities, and

has led to debate within the academic community on how

best to ensure the transfer of new ideas and technology from

the laboratory to the wider society.

The opportunities generated by industry's increasing

support of research are manifold. They include the prospect

of stable, long-term funding which may complement govern

ment support of basic research, which has declined in real

terms in recent years. In addition, and importantly, both

universities and industry can benefit from closer com

munication and ties. After all, both rely, for their own evolu

tion and growth, on the talent and new ideas generated by a

vigorous system of higher education. Further, new ideas and

technologies which are born in an academic setting must be

nurtured and developed before they lead to broad practical

uses. This development usually occurs in a business setting,

and is aided by effective communication between universities

and business. Finally, universities can better chart their own

future development in certain areas, such as engineering, if

they know about the changing directions and needs in the

world of business and industry. This does not mean that

universities will forsake their traditional independence and

reliance on their faculties to pursue shifting frontiers of

knowledge. It does mean, however, that our development

can, and should, be informed by these perspectives.

At the same time that we recognize the opportunities

and mutual advantages of closer ties between academia and

industry, we must consider the questions arising from such

association.

The sponsorship of research by business and industry

has, at times, created considerable anxiety within univer

sities, and has been the subject of considerable public interest

and commentary in the media as well. This is not surprising,

because such collaboration carries the possibility of conflict

between the essential openness and public accountability of

the universities, on the one hand, and the private and pro

prietary interests of industry on the other. In a context in

which knowledge becomes not only power but wealth, ques

tions arise about intellectual property rights, about the clos

ing down of communications between research colleagues,

and about the effect of varying degrees of openness or

secrecy on the progress and integrity of research. For some,

industrially sponsored research appears to increase the prob

ability that universities will be compromised in their in

dependence and objectivity, or that their priorities and ac

tivities will be distorted by private influence.

These kinds of questions led to a gathering last March

of people from five major research universities (the Califor

nia Institute of Technology, Harvard University, MIT, Stan

ford University, and the University of California) at Pajaro

Dunes, California, to explore the issues generated by com

mercial sponsorship of research and other forms of interac

tion between industry and academe. The meeting was

A20 JANUARY 1983

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Ccn£uJ;ntj ana Contract

R i D m Antennas. Preptnahen Microwave Sys

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avail a bio

George Ploussics

President. '55

PC B«x26BVI.STAHackney Circle

Anlov4*r. MA 018IC

'617)475-037:*

. Contract Research andDevelopment and

Consulting in RadioFrequency and MicrowaveEngirveenng and RelatedAreas

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01801

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Consulting EngineersRail Transportation

Thomas K Dyer '43

1762 Massachusetts AtoLexington. MA 02173

(617)662-2075

Washington, DC(202) 466-7755

Chicago. IL(312)663-1575

Philadelphia. PA

(215)569-1795

\

Entjinoerjig Ccrisut&ng

Services and R & DnEnergy and Power

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R & D tn Energy EfbnentEquipment and Processes

E S, Mteras '56

238 Main St., Kendall SqCambrtdge, MA 02142

(617) 492-3700

Puzzle Corner

Allan J. Gottlieb

Dropping a Rock in

Borneo

ft! '■■

Allan J.Gottlieb, '67, is at

the Courant Institute ol

Mathematical Sciences,

New York University. He

studied mathematics at

M.I.T. and Brandeis.

Send problems,

solutions, and comments

to him at the Courant

Institute, New York

University, 251 Mercer

St., New York, N.Y.,

10012.

Not very much to report this month. I am

very busy helping raise our son David

and managing a research group in "paral

lel processing" (using many cooperating

computers to solve large problems much

faster than is possible with a single

machine). I find both parenthood and

computers to be extremely interesting

and rewarding; I just wish every day were

a little longer so I could spend more time

on each.

Problems

F/M 1 Our first problem for this month is

from Bob Kimble and Jacques Labelle,

who ask that White mate in three:

F/M 2 Next we have a geometry problem

from Greg Huber: Let T be the torus ob

tained by rotating the circle (x - 2) + y =

1 in the xy-plane about the y-axis. Let P

be a plane tangent to the torus T at the

point (0, 0, 1). Find the volume and sur

face area of the small region obtained by

slicing T with P.

F/M 3 Norman Wickstrand wants you to

solve the simultaneous equations:

x°y = (y* + 1)x3

y°x =9(x2 + Ity*

F/M 4 John Prussing adds a new wrinkle

to an old problem:

A pilot flies south over a spherical earth a

distance D, flies due east a distance D,

and then flies due north a distance D, ar

riving back at precisely the starting point.

For D equal to the radius of the earth, find

all solutions for the starting latitude.

F/M 5 Bruce Calder hung a vertical

plumb line down the center of a one-

mile-deep mine shaft somewhere in Bor

neo. He dropped a rock from right next to

the top of the plumb line. How far from

the bottom of the plumb line will the rock

land if we ignore friction?

Speed Department

F/M SD 1 Smith Turner sailed a 10-by-

30-foot boat into a canal lock and then

threw overboard a 10-cubic-foot chest

having a specific gravity of 4. What hap

pened to the water level in the lock? -

F/M SD 2 A bridge quickie from Doug

Van Patter:

North:

A K

7¥

♦

6

J 8

5

KQJ7

K

K

A

South:

105 4

A Q 10 4

10 7

8 6 3

Your partner (North) opened one club

and raised your one-heart bid to four. The

opening lead is 4Q. What play gives you

the best chance of making your contract?

Solutions

OCT 1 South is on lead with hearts as trump and is

to take all tricks against the best defense:

♦ A J

V 5♦ K Q

♦ 108

A 43♦ K2

¥ —♦ J 98

♦ 97 65

4k Q

8 4

« A 1032

♦ -The following solution is from Matt BenDaniel. There

are three considerations:

D We must pull East's trumps; this can only be done

by finessing from the dummy.

D The bad offensive lit combined with East's

diamond singleton limit transportation (or the offense.

D West must discard the protection from his poten

tial winners.

The basic plan, then, is to get to board in spades, fi

nesse East once, arid play another round of hearts.This will pull East's trumps and will also cause West

to discard. West's discards will determine the de

clarer's play for the rest of the hand. The following

diagram gives the details of play:

A22 FEBRUARY/MARCH IV8J

with implememing M.l.T.'s commitment to research

and teaching in health economics and policy. Funds

Irom the Henry J. Kaiser Family Foundation will be

available lor student support.

Dorothy Eleanor Westney, who arrived at

M.I.T. to be assistant professor in international

management last fall, is now the first Mitsubishi

Career Development Professor. The chair is intended for junior faculty with a strong interest in

comparative management who will devote at least

a portion of their research and teaching to better

understanding Japanese culture and Institutions.

Dr. Westney, who is fluent in Japanese, holdsdegrees in sociology from the University of To

ronto and Princeton and for four years held a joint

appointment in sociology and mangement at Yale.

David L. Bodde, S.M.73, is currently assistant

director. Congressional Budget Office. . ..William M. Nuckols, S.M.'64, has been named

vice-president for business planning and analysis

at the Penn Central Corp.. New York City. ...

Bernard J. Jourdan, S.M.70. writes, "I am back

in the states as executive vice-president for the

Campagnie Generale des Cieux U.S. Group. Imoved to Philadelphia, Penn., from Paris in

January 1982. I married In late 1981 and my wife

is expecting a baby." . . . Gordon C. Shaw,

DS.M.'eo, professor of administrative studies atYork University, Canada, has just completed a

two-year study to forecast the number of

Canadian-registered dry-bulk vessels required toserve Great Lakes cargoes in 1990. The final report has been published by the University of

Toronto/York University Joint Program in Transportation.

The following graduates participated in theM.I.T. Alumni Fund upgrading telethon: Samuel

Appleton, S.M/57; Carol Bratley; Peter Con-

dakes, S.M.'SO: Stephen Hall, S.M. 62; Howard

Hlllman, S.M. 60; Walter Lehmann, S.M.'75;

Howard Miller, S.M.'63 (coordinator and top cal

ler); Constance Stubbs, S.M.79; and ThomasThompson, S.M.'57.

Sloan Fellows

Donald H. White, S.M.70, senior vice-presidentat Hughes Aircraft Co., Culver City, Calif.,- has

been named a director of the company.... Hugh

E. Witt, S.M.'57, vice-president, government

liaison, United Technologies Corp., has been

elected vice-chairman ol Aerospace Industries

Association of America, Inc.. Washington, D.C.

... Claudia B. Llebesny, S.M/80. is currently

product manager—fused and sintered specialty

products at the Norton Co.. Worcester, Mass.

Jere Drummond, S.M.77, is currently vice-

president of Southern Bell Telephone & Telegraph

Co., Charoltte, N.C C. Clement Patton, 77,

is currently vice-president of Southern Bell Telephone & Telegraph Co., Atlanta, Ga.

Senior Executives

Alexander M. Williams, '63, former

president—U.S. Division of the Campbell SoupCo., Camden, N.J., is currently president-

International Division. ... Joseph A. Baute, '64,

chairman and chief executive officer of the Mar-

kern Corp., has recently been named a director ol

Houghton Mifflin Co., Boston, Mass.

Julian Hartwell, '67, of Cohasset, Mass., and a

long-time executive with New England TelephoneCo., passed away on September 25, 1982.

XVIAeronautics and Astronautics

What happened to the Navy's fleet defense mis

sile system known as Talos? It was the first

ramjet-powered guided missile, first conceived in

194S and delivered to the Navy in 19SS. But it was

never used to its full capability, say two

analysts—Rear Admiral Wayne E. Meyer, '47,

and Captain (Ret.) Richard W. Anderson,

S.M.'60—in the spring of 1982 issue of John Hop

kins Physics Laboratory's Technical Digest.

Talos' electronics became obsolete before their

time, maintenance was high, and detection sys

tems were inadequate to locate targets at a dis

tance large enough for Talos to be fully effective,

write authors Meyer and Anderson. Among the

reasons is that Talos never had an at-sea devel

opment site except on operational ships of the

fleet; White Sands Missile Flange was the only

test facility that could handle Talos.

Professor Emeritus C. Stark Draper, '26, re

ceived the Medal of the City of Paris as guest ofhonor at the 33rd Congress of the International

Astronautics Federation in Paris late last summer.

During the Congress "Doc" Draper resigned his

post as president of the lAFs Academy of As

tronautics after 19 years of service. Preceding his

engagements in Paris, "Doc" had attended the

UNISPACE '82 Conference in Vienna and the His

tory of Astronautics and Rocketry Conference in

Moscow, the latter to celebrate the 25th anniver

sary of Cosmonaut Yuri Gagarin's first manned

spaceflight. "Doc" was joined in Paris by Janet B.

Jones-Ollvorlo, a graduate student in the de

partment at M.I.T., who received the lAF's Ed-

mond A. Brun Medal for her paper on the design of

an environmental research facility for low earth

orbit.

Sir William R. Hawthorne, Sc.D/39, master of

Churchill College, Oxford, who is senior lecturer

at M.I.T., delivered the Calvin W. Rice Lecture at

the 1982 winter meeting of the American Society

of Mechanical Engineers and at the same meetingwas designated an honorary member of ASME.

The Rice Lecture topic: "World Energy versus the

Environment—Conflict or Compromise."

James A. Martin, S.M.'69, has recently com

pleted (February 1982) his D.Sc. degree from

George Washington University. ... Leslie M.

(Bud) Boring, S.M. 64, reports, "Not too much in

the way of news this past year, although the con

struction and engineering contractor banking

business in France has gone well, the change in

government notwithstanding, and our two children

(girl 2 and boy 6) continue to grow marvelously.

On the avocation front, have made good progress

with my painting, which I started at evening

classes at the Musuem of Fine Arts while still at

M.I.T. (now beginning to be decades ago!). I was

accepted for exhibition at the Grand Palais in

Paris for both the Salon d'Automore and Salon

des Artistes Francais and even won a medal in our

hometown exhibition, the Prix de la Villa de

Croissy-sur-Seine! Have been contacted by sev

eral galleries, although their initial demonstration

of enthusiasm drops to zero when'they learn I'm a

banker and do not paint full time. It seems volume

of productoin is what ultimately counts. Anyway,

it's a lot of fun and doubly interesting as we live

within 20 minutes' walk of several scenes painted

by the Impressionists."

James Harrlll, S.M.'64; Gaylord MacCartney,

S.M. 53; and Robert Stern, Sc.D.'63, participated

in the M.I.T. Alumni Fund fall upgrading telethon.

William B. Abbott Ill, S.M/61, reports that he

has retired from the U.S. Navy after 30 years of

service. For the past 19 years he served in the

Strategic Systems Project Office where he was

Navy manager of the Polaris, Poseidon, and Tri

dent submarine-launched ballistic missile pro

grams.

XVIIPolitical Science

A new curriculum leading to a master's degree in

political science and public policy has been ap

proved by the faculty, and the first students will

enter next fall. The program differs from the regu

lar master's program in political science in its em

phasis on policy studies in one of four fields: de

fense and arms control; science, technology, and

public policy; communications; and international

development. Mid-career professionals as well as

recent bachelor's graduates are welcome as ap

plicants; they will finish the program in one to two

years, depending on preparation, and wilt be

qualified for further studies toward the Ph.D. and

for positions in government, business, and non

profit institutions. Professor Donald L. M. Black-

mer assured the faculty that the department is

"committed to the training of policy analysts with a

broad outlook. Technical, analytical skills are im

portant," he said, "but we believe that policy ana

lysts must also have a larger understanding of

what they are doing—of long- as well as short-

term consequences of public intervention, of unin

tended as well as intended consequences, and of

wider as well as narrower effects of policies."

XVIIIMathematics

Professor Daniel M. Kan, a member of the M.I.T.

faculty since 19S9, has been honored by election

to the Royal Netherlands Academy of Arts and

Sciences; a native of Holland, Prolesscr Kan'sbachelor's and master's degrees are from the

University of Amsterdam.

Jerry Grossman, Ph.D.74, reports that he is

currently ^associate professor of mathematical

sciences at Oakland University, Rochester, Mich.,

and assistant department chairman. He was mar

ried to Suzanne Zeitman on August 15,, 1982....

Margaret Freeman, S.M.34, was a caller in theM.I.T. Alumni Fund fall upgrading telethon.

XIXMeteorology

George F. Collins, S.M.'48, has joined the En

vironmental Resource Planning Division of

Charles T. Main, Inc., Boston, Mass., as head of

air sciences. ... Ryland Y. Bailey, 52, is cur

rently senior engineer of the State Corporation

Commission ol Virginia and a member of theMethodist Church, Richmond.

XXIHumanities

Carl Kaysen, David W. Skinner Professor of Polit

ical Economy who is director of the Program in

Science. Technology, and Society, is a member of

a committee of the American Academy of Arts and

Sciences to oversee the academy's new associa

tion with the International Institute for Applied

Systems Analysis, near Vienna. American supportfor I IASA had previously been organized through

the National Academy of Sciences in Washington,but political pressure stemming from White House

concern about security and possible technology

transfer to the East caused NAS to withdraw.

Science, Technology, and Human Values, aquarterly edited by Marcel La Follette in the

Program in Science, Technology, and Society and

cosponsored by Harvard's Kennedy School of

Government, is now being published by John

Wiley and Sons, Inc., New York; it has previously

been in the periodicals group of the M.I.T. Press.

Technology and Policy Program

Barbara Herrmann, S.M.'81, is currently vice-

president of Consulting Resources Corp., a man

agement consulting firm which specializes in serv

ing chemical process industries. Her firm has re

cently been recognized nationally for its work in

commodity and specialty chemical markets. ...

Tarlq Mahmood, S.M.'SO, has accepted a new

position with the Power Rates Section of the De

partment of Public Service, Albany, N.Y.—

Professor Richard de Neulville, '60, Chairman,

Room 1-138. M.I.T., Cambridge, MA 02139.

TECHNOLOGY REVIEW A2t

Also solved by Gian Kolderness, John Woolston,

Matthew Fountain, Doug Van Patter, Matt Daniel,

Winslow Hartford, John Boynton, and the proposer,

Emmet Duffy.

OCT 2 Solve the set of four cryptarithmelic puzzles,

entitled "Seven and Twelve." created by Nobuyuki

Yoshigahara.

Only Harry Hazzard was able to solve this one and

even he could not find a solution to the third puzzle.

13677

47870

47870

47870

157287

2:

107

47570

47570

47570

47570

47570

237957A-

8162

8162

42623

42623

42623

42623

202623

202623

13788

48280

48280

48280

156628

107

87570

87570

87570

87570

87570

437957

17566

36460

36460

36460

126946

129

69892

69892

69892

69892

69892

349569

37166

96460

96460

96460

326546

567

27476

27476

27476

27476

27476

137947

567

37476

37476

37476

37476

37476

187947

592062

OCT 3 Given an ice cream cone filled with water,

how large a sphere displaces the most water? Let thehalf angle of the cone be x, and let the radius of thebase be 2.

Emmet Duffy sent us a fine solution of his own aswell as a reference to a similar problem in the classic1904-calculus book of Granville et al. Leo Harton

used the well known symbolic algebra package

MACSYMA and obtained an "imaginary" sphere

(see the end of the following solution). Mr. Duffy

writes:

Assume the sphere is tangent to the side of the

cone. Let b = depth of cone and a - distance from

center of sphere to bottom. Then 2/b = tan x and b =

2/tan x, or b = (2 cos x)/sin x; r/a = sin x and a = r/sin

x. Then

h = b-a = (2cosx- r)/sin x.

The volume, v, of displaced water is given by:

v = v (2rJ/3 + r'h - hV3), where h is not greater

than + r nor less than - r.

v = it (2rV3 + 2r» cos x/sin x) - rVsin x - (2 cos x -

r)»/3 sin3.

dv/dr = 7t |2r' + 4rcosx/sinx - 3r*/sinx - (2cosx -

Oi-IJ/sin'x]

dv/dr = jt [2r» + 4rcosx/sinx - SrVsinx + (4cos'x -

4rcosx + r'J/stn'x]

Set dv/dr to zero, clear fractions, and divide by n.Then:

2rfsin*x + 4rcosxsin'x - 3r"sin»x + 4cos'x -

4rcosx + r1 = 0

Collecting terms:

r^sin'x - 3sin'x + 1) + r(4cosxsin"x - 4cosx) +4cos»x = 0.

r'(2sinax - 3sin'x + 1) + 4rcosx(sin*x - 1) + 4(1 -

sin'x) = 0.

Factoring:

r'(2sinx + 1)(sinx - 1)(sinx - 1) + 4rcosx(sinx +

1)(slnx - 1) +4(1 + sinx)(1 - sinx) = 0.

Divide by (sinx - 1):

r"(2sinx + 1)(sinx - 1) + 4rcosx(sinx + 1) - 4(stnx

+ 1) =0.

Solving this using the quadratic formula yields theexpressions for r shown at the bottom of this column:

the desired value of r is

r = - 4cosx/2(2sinx + 1)(sinx - 1) = 2cosx/(2slnx

+ 1 )(1 - sinx) which converts to:

r = 2cosx/(sinx + 1 - 2sin*x) or

2cosx/(sinx + cos2x).

The other value of r is

r = (- 8cosxsinx - 4cosxV2(2sinx + 1)(sinx - 1).

This simplifies to r = 2cosx/(1 - sinx), the radius of

an imaginary sphere tangent to the imaginary exten

sion of the cone and also tangent to the tip of the

water level of a full glass, displacing no water.

Also solved by Winslow Hartford, Norman

Wickstrand, Matthew Fountain, David Evans, Harry

Zaremba. William Moody, and the proposer, EdmundNadler.

OCT 4 Find the closed form (not infinite series) solution of dy/dx = x -y1.

The following solution is from John Wrench:

The differential equation dy/dx = x - y1 is a special

case of Recall's equatbn and is transformable to a

r = {- 4cosx(sinx + 1) ± [16coslx(sinx + 1)» + 16(2sinx + 1)(sin"x - 1)|t)/2(2sinx + 1)(sinx -r = {- 4cosx(sinx + 1) + [16cos»x(sinx + 1)» - 16cos'x(2sinx + t)JI)/2(2sinx + 1)(sinx - 1)r = (- 4cosx(sinx + 1) ± (16cosIxsln»x)tJ/2(2sinx + 1)(sinx - 1)r = [- 4cosx(sinx + 1) ± 4cosxsinxl/2(2sinx + 1)(sinx - 1)

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second-order linear homogeneous differential equa

tion (called Airy's equation):

d»u/dx'-xu =0.

by the substitution y = 1/u diVdx. The general solu-

tbn of the transformed equation is

u = el xJ llo(2ra x'-») + c2 x J l.in(2/3 x'•»).

where the Is are modified Bessel functions of the first

kind of orders 1/3 and -1/3. If we use the differential

formulas

d/dx |u» lp(u)| = u» lp.,(u)du/dx, and

d/dx |u-» yu)! = u-» t.,(u)du/dx,

we obtain

diVdx = d x l-,n(2/3 x'J) + c2 x lt3<2/3 x1J).

Thus, the general solution of the given differentialequation may be written

y ell ,,,(2/3 x >•*) + c2 I .,,,(2/3 x >•»)

Clearly, that solution involves only one independent

constant—that is, d/c2 or its reciprocal.

Also solved by Leo Harten. Antony Beris, Matthew

Fountain, and David Evans.

OCT 5 Given a regular hexagon with an inscribed

circle, whal is the ratio of the area of the six smaller

circles (see drawing) to the area of the inscribed cir

cle?

Roger Milkman had little trouble with this one:

The answer is two-thirds, assuming that the rate de

sired is that of the total area of the six small circles to

the area of the large one. Set the radius of the in

scribed large circle at 1. Draw it vertically upward

from the center; it is perpendicular to, and bisects,

the top side of the hexagon. Each side of each of the

equilateral triangles is therefore sec 30° (since the

radius so drawn also bisects the lowest angle of its

triangle). The radius of (for example) the upper left

hand small circle will be perpendicular to, and bisect,

any side of its triangle that it reaches. It will also

bisect the opposite angle if prolonged. The radius of

the small circle is thus [Yi sec 30°) tan 30°, or 1/3.

The ratio of the areas is.6 times the ratio of the

squared radii, or 6/9, or 2/3.

Also solved by John Woolston. Emmet Duffy,

David Evans, Norman Wickstrand, Winslow Hartford,

Naomi Markowitz, Irl Smith, Mary Lindenberg,

Matthew Fountain, Jordan Wouk, Steve Feldman,

David Lukens, Phelp Meaker, and Raymond Gail-

lard.

Better Late Than Never

M/A 4. Robert Kennedy has responded.

JUL 1, JUL 3, and JUL 5. Richard Kess has re

sponded.

A/S S. John Woolston has responded.

Y19B2. Harry Hazard has responded.

OCT SD1. The proposer doesn't feel the problem is

speedy.

OCT SD2. The proposer believes that calculating

the last eight digits is more fun.

Proposers' Solutions to Speed Problems

F/M SD 1 The level drops 1.2 inches.

F/M SD 2 The *5. West may continue a second

spade instead of shifting to a diamond honor.

A24 FEBRUARY/MARCH 1983

Puzzle Corner

Allan J. Gottlieb

Sneaky vs. Freddy

Allan J. Gottlieb. 67, is

associate research

professor at the Courent

Institute of Mathematical

Sciences of New York

University; he studied

mathematics at M.I.T.

and Brandeis. Send

problems, solutions, and

comments to him at the

Courant Institute, New

York University, 251

Mercer St., New York',N.Y., 10012.

Since I've not reviewed the ground rulesof "Puzzle Corner" for more than a year,let me do so now for the benefit of newreaders.

In each issue we present five regular

problems (the first of which is chess- orbridge-related) and two "speed" problems. Readers are invited to submit so

lutions to the regular problems, and onesubmitted solution for each problem is

selected for publication three issueslater. We also list other readers whose

solutions were successful. For example,solutions to the problems you see below

will appear in the August/September is

sue. Since I must submit that column

sometime in May (today is January 15),

you should send your solutions to me dur

ing the next few weeks. Late solutions, as

well as comments on published solutions, are acknowledged in subsequentissues in the section entitled "Better Latethan Never." For example, commentsappear in this issue on previously pub

lished solutions to two problems, JUL 1and Y1982.

For "speed" problems the procedure isquite different. Often whimsical, theseproblems should not be taken too se

riously. If the proposer submits a solution

with the problem, that solution appears at

the end of the same column in which theproblem is published. For example, so

lutions to the "speed" problems printed inthis issue are given at the end of this column. Only rarely are comments on

"speed" problems published or acknowledged.

There is also an annual problem, published in the first issue of each new year;

and sometimes we go back into history to

republish problems which remained un- 'solved after their first appearance.

All problems come from readers, and

all readers are invited to submit favorites.

I'll report on the size of the backlog, and

on criteria used in selecting problems for

publication, in future issues.

Problems

APR 1 Here is a zany problem from Ar

thur Polansky, who recounts a conversation with a friend:

"I know you've played duplicate bridgewith life masters and once received an

invitation to the Brisbane regionals. But

have you ever played mixed-pairs quad

ruplicate?" my friend asked. Reluctantly Iconfessed that I had not. My friend ex

plained: "Each team of four players is

split amongst four tables so that one

member plays North, one South, one

East, and one West. Each deal of thecards is played four times, once at eachtable."

"But that means that I always get apartner from an opposing team," I com

plained.

"Precisely," my friend said. "Cuts

down on preplanned conventions, likethe 17-19-point left-eye-scratch. Also

does wonders for the scoring." I could

imagine. "Consider deal 23 from last

night. Team 5, the Four Spades, reachedfour spades at all four tables."

By now I was mildly disgusted. "Withsuch bidding prowess, why do you allowthese turkeys to keep playing?"

"They own the club," was my friend'sanswer.

Hastily bidding adieu (which was im

mediately doubled for takeout), I raced

home. Then I realized that the self-styledtournament director had failed to show

me the deal in questbn. Can you? (That

is, find a deal such that four spades,when played from any position, will be set

at least seven tricks (i.e., the defense will

make four spades). Naturally, the set willbe perforce: the declarer will do everything in his power to prevent being set

and to minimize the damages if being setbecomes inevitable.)

APR 2 Frank Rubin wants a defect-freeplane of words:

Shown below is a simple word square.ACE

PAR

ERA

We could fill the plane with such word

squares so that every cell of an infinite

grid lay at the intersection of two words.However, such an arrangement wouldhave two defects: (1) the pattern would

be repetitive, and (2) the words would no

all be interconnected. Show how thes<

defects can be fixed; i.e., fill the plantwith an infinite number of words so tha

the pattern does not repeat in any row o

column, every cell lies at the intersectioi

of two words, and every word is con

nected to every other one by a chain o

intersecting words. Do not use any twoletter words.

APR 3 Here's one from Ming Chung

who first published it (and the diagram a

the bottom of this page) in the studen

newsletter of the M.I.T. Physics De

partment:

One afternoon in a swamp Sneaky th€

snake, whose mass is Ms and length t%

was resting on the left end of a log o

length L and mass M,, when he suddenly

remembered that he was hungry. After s

few moments of looking around, he founcFreddy the frog sitting on a large disk-

shaped leaf of radius R and mass M,, h

cm. away from the center of the disk. Ini

tially the disk was 4/5 ts cm. away from

the right end of the log. Quite naturallySneaky started moving toward Freddy,

and Freddy, who was ignorant of the laws

of physics, started jumping for his life.However, when Freddy remembered thatSneaky had his best friend for breakfast

in the morning he was overcome by therage and desire for revenge, and he

turned around to charge against Sneaky.

Well, normally it would have been a

dumb move. But the laws of physics re

ward the courageous; when Freddystopped at a point on the disk and

realized that his attempt was futile if not

suicidal, the disk was just out of the reach

of Sneaky, who was on the right end ofthe log by that time. Sneaky could not

swim, so the good guy was saved and

the bad guy had a hungry afternoon. It's adumb story, you say. Well, if you'd like to

entertain your intelligence, consider the

following question: Can you locate the

point where Freddy stopped? Assume allmotions occurred on a line, and the water

offered no resistance. Both the log anddisk have a uniform mass density.

APR 4 Ronald Burde doesn't seem to

know enough to come in out of the rain.Instead he asks:

Assuming the sudden onset of steady

rainfall, will a person remain dryer .by

walking or running any given distance to

shelter? (Mr. Burde remarks that this

problem has reached the level of a

friendly "cause celebre" among some

R

A22 APRIL I»M

ous yacht builders and owns his own company

building traditional cruising sailing yachts in

molded fiberglass. ... Donald Distant, S.M.'SO,

has been appointed chief engineer (electrical andmarine) with the-Port of Singapore Authority in

May 1982, responsible for all electrical engineering and marine engineering activities in P.S.A.

Theotokis S. Mllas, S.M.'7t, is presently

superintendant engineer with Cove Shipping, New

York City. ... Robert J. Bosnak, '60, chief,

Mechanical Engineering Branch, Division of En

gineering of the U.S. Nuclear Regulatory Com

mission, was a recipient of the 1982 American

Society of Mechanical Engineers' Bernard F.

Langer Nuclear Codes and Standards Award. Theaward was for "outstanding professional and

technical contributions in the development of

ASME nuclear codes, standards, and accredita

tion programs." ... Robert I. Price, 'S3. recently

retired as the third ranking officer of the U.S.

Coast Guard, commanding both the New York

District and the Atlantic area, has received the

annual Vice Admiral "Jerry" Lane Medal for "out

standing accomplishment in the marine field," by

the Society of Naval Architects and Marine En

gineers. His naval career has focused on the im

provement of national and International standards

for maritime safety and environmental protection.

XIVEconomics

Arthur G. Ashbrook, Jr., Ph.D.'47, reports that

he has retired from the Central Intelligence

Agency after 28 years of service as an economist.

For four years he was assigned to the faculty at

the National War College. Washington, D.C., and

intends to continue teaching and research in the

"dismal science." ... Two graduates of the de

partment have been re-elected to Congress:

Howard E. Wolpe, Ph.D/62. won his third con

secutive term in Congress from Michigan's third

Congressional district, the area encompassing

Kalamazop; and Les Aapln, Ph.D.'66, won his

seventh consecutive term in Congress from Wis

consin's first district.

XVManagement

Project management will be so different by 1990

that "a new breed of manager will be needed to

cope with changes." says Albert J. Kelley, '48,

president of the Arthur D. Little Program System

Management Co. The changes: increasing con

straints by such external factors as environmental

and government regulation, multiple financial ar

rangements, new risk due to advanced, unproved

technology, and inflationary pressures. These is

sues and others are the subject of New Dimen

sions of Protect Management (Lexington Books,

1983), a book of 15 essays by Dr. Kelley and other

project management specialists including John

R. White, S.M/63, senior vice-president-

strategic planning at Arthur D. Little. Inc.; Profes

sor Mel Horwlch of the Sloan School of Manage

ment; John F. Magee, president of Arthur D. Lit

tle, Inc.; and Robert C. Seamans, Jr., Sc.D/51,

Henry R. Luce Professor of Environment and

Public Policy at M.l.T.

Richard A. Mlchaelson, S.M.77, reports, "I

have recently been made vice-president at Met-

Path, Inc., responsible for billing and receivables.

MetPath, a wholly-owned subsidary of Corning

Glass, is in the clinical laboratory business. ...Warren H. Hausman, Ph.D.'66, has been ap

pointed chairman of the Industrial Engineering

and Engineering Management Department at

Stanford Univeristy, Palo Alto, Calif. His area of

expertise is in production/operations manage

ment and analysis. . . . George E. Williams,

S.M'49. corporate vice-president of United

Technologies Corp., has taken an early retirement

and has joined Kensington Management Consul

tants, Stamford, Conn., as a senior vice-

president.

Charles C. Holcomb, S.M.75. is currently the

commanding officer of USS Campus and resides

in Charleston. S.C Arlstea Xafa, S.M.75. has

left her London-based position as vice-

president—finance of Saudi REDEC and is now in

charge of Boston-based Global Investments Lim

ited. She writes that she would like to hear from

alumni in microelectronics and computer man

ufacturing concerns and those engaged in R&D

equity financing. ... W. John Swartz, S.M.'67,

has been appointed executive vice-president of

Santa Fe Industries. Inc., Chicago, III William

O. Schach, S.M.'5O, is currently senior vice-

president of Merrill Lynch, Pierce, Fenner, and

Smith. Inc.

Frederic C. Westendorf, S.M.64. is currently

business planning manager for IBM Europe/ Mid

dle East/Africa. White Plains. N.Y. ... Debra

Greenberg, S.M.78, reports that she has

founded her own consulting firm in New York City.

Otto H. Poensgen, Ph.D.'64, who was assis

tant professor at M.l.T.'s Sloan School (upon his

graduation until 1967) and later was appointed to

a chaired professorship in industrial management

at the University of'Saarland, Saarbrucken, Ger

many, passed away oh October 29, 1982. He waswell known in the U.S. and abroad for his research

and teaching in management science, business

strategy, and organizational structure. During a

sabbatical leave in 1980, he continued his re

search at M.l.T. ... Donald G. Robbins, Jr., of

Fairfield. Conn., passed away on October 11,

1982: no details are available.

Sloan Fellows

Ormand J. Wade, S.M.73, president and chief

operating officer of Illinois Bell Telephone Co.,

has been elected to the boards of Harris

Bankcorp, Inc., and Harris Trust and Savings

Bank Leroy E. Day, S.M'60, writes. "I retired

from NASA and the Space Shuttle Program to

start my own consulting business for aerospace

and management—business expanding. I was

also awarded NASA's Distinguished Service

Medal for work on the space shuttle." ... Brian J.

Kelly, S.M.73. vice-president of marketing of Bell

of Pennslyvania and Diamond Star Telephone

Co., has been elected to the Board of Trustees of

Hahnemann University. Philadelphia, Penn.

John C. Davis, S.M.'56. senior vice-president

and a director of Santa Fe Railway. Chicago, re

tired from the firm in December 1982. ... Eric

W.A. Lange, S.M.'62. has joined the staff of Fail

ure Analysis Associates, Troy, Mich. ... Sam R.

Willcoxon, S.M/65, formerly vice-president of

American Telephone & Telegraph Co., New York

City, has become the firm's executive vice-

president of marketing (interexchange organiza

tion) William H. Springer, S.M/68, executive

vice-president of finance for Illinois Bell Tele

phone Co., has become its senior vice-president

and secretary.

Steven J. Miller, S.M.79. writes. "I was re

cently married to Janet Patricia Beal on Sep

tember 11. 1982. and am currently director of

marketing—western region tor" Data Resources.

Inc.. a leader in economic forecasting and consult

ing." . .. William S. Wheeler, Jr., S.M.'54. re

ports. "I look early retirement as a senior vice-

president of Englehard Industries earlier this year

and with an associate bought a small manufactur

ing company which manufactures bulk material

handling equipment. The company is Buck El, Inc.

Thoroughly enjoying the independence of being

the C.E.O. of my own company." . .. Cllne W.

Frasler, S.M.72, a senior technical member of

the Manufacturing Automation and Computation

Department at the Charles Stark Draper Labora

tory. Cambridge, has been appointed head of the

department.

Robert E. Workman, S.M.'SS, a retiree after 39

years of service at Goodyear Tire and Rubber

Co.,- Akron, Ohio (his latest position was vice-

president of general products development),

passed away on November 18, 1982. He was a

past president of the'International Institute of

Synthetic Rubber Producers, a member of the

American Chemical Society and of the American

Institute of Chemical Engineers.

Management and Technology

Program

Geoffrey N. Andrews, S.M.'82, is currently with

the New Opportunities Department at Pilklngton

Brothers in England, where he recently headed acorporate-wide project called Quest—ferreting

out underutilized skills and technology within the

firm. He saw Professor Edward B. Roberts,

Ph.D.'57. briefly during Ed's trip to Pilkington at

the end of November.

XVIAeronautics and Astronautics

The spread of nuclear power to developing na

tions is dangerous because it lowers their

threshold for nuclear weapons development, says

John P. Holdren, '65, professor of energy and

resources at the University of California at Berke

ley. A nuclear power program assures the pres

ence of the technical skills needed to operate a

weapons program and often includes facilities for

coverting reactor fuel into weapons-usable mate

rial. Dr. Holdren wrote early this year in the Bulle

tin ol the Atomic Scientists. A power program also

reduces the marginal cost of a weapons program,

and it lowers some of the political barriers to nu

clear weapons commitments. Dr. Holdren said.

Roger Neeland, S.M.70, writes, "After teach

ing at the U.S. Air Force Academy for several

years in the Department of Astronautics, and two

years heading FAA's Airborne Systems Branch in

their Systems Research and Development Ser

vice, I am now responsible for analysis to support

the Air Force Systems Command long-range

planning." .. . Louis H. Benzlng, S.M.'S2,

executive vice-president of Jofee Marketing Inter

national, has taken on the additional position of

president and chief executive officer at the Micro

Z Corp., Monrovia. Calif Tore Christiansen,

S.M.°82, has recently taken up an engineering

position in Det Norske Veritas. an international

Norwegian classification society for ships and

off-shore structures, and is presently working in

Oslo, Norway.

Henry F. Lloyd, S.M/46. reports that he is still

a college administrator at Flagler College and has

been in this position since retirement from the

U.S. Navy. ... Robert Schogerln, S.M.74. is

presently employed at Talasa Law and Technol

ogies, Woodbury, Conn. ... Jack E. Stelner,

S.M/41, vice-president of corporate product de

velopment at the Boeing Co., recently served as a

witness on hearings on government policy toward

aeronautical research chaired by Congressman

Dan Glickman, chairman of the Subcommittee on

Transportation, Aviation and Materials.

Technology and Policy Program

Julian Vlllalba, S.M.'SO, has recently completed

his doctorate at M.l.T. in international technology

transfer. He will teach at the Institute des Estudios

Superlores y Administracion, Caracas, and also

plans to do consulting work on state-owned en

terprises and public administration. ... David

Rubin, S.M.'BI is working for the city of San

Francisco as a member of a task force studying

district heating systems.—Richard de Neufville,

Chairman, Technology and Policy Program,

M.l.T., Room 1-136, Cambridge, MA 02139.

TECHNOLOGY REVIEW A2I

fellow-alumni. He agrees with them that

the solution to the problem is straightforward and that it has been solved many

times, but agreement is lacking on theanswer.)

APR 5 Peter Groot wants you to find

positive integers a, b, and c such that

a3 + b< = c5.

Speed Department

APR SD1 Rufus Briggs will let you have

additional information, but not much:

Given a glass tube 6 inches in diameter

and 60 inches long. The tube is cappedand sealed at one end and mounted ver

tically. The bottom 30 inches of the tube

is filled with thick molasses and the top30 inches with a medium grade of oil. A

1-inch polished steel ball is allowed to fall

through the liquids, and it takes t0 = 2.73seconds to fall half way down the tube.

What is the time tT for the ball to fall the

entire length of the tube? To answer thisquestion you may have one and only one

additional piece of information. Identifythe information you have chosen, and

write the equation for solving the problem.

APR SD2 David Evans wonders whatcommon five-letter word is spelled wrong

by nearly every M.I.T. graduate.

Solutions

HID 1 White to mate in three moves:

H

Avl Ornstein found that analytic geometry

makes this problem rather easy: Let AC and BQ

be the x and y coordinates, respectively. In addition, let A be (-a, 0), let B be (0. b), and let C be

(c, 0). ThenEis(-a -b, a), Fis(b +c,c),andQ

is (0, - a - c). The slope of AF is c/(a + b + c)and Its y-lntercept Is ac/(a + b + c). The slope of

EC is -a/(a + b + c) and Its y-intercept Is ac/(a +

b -i- c). AF and EC therefore meet on line BQ atpoint Z.

Also solved by Winslow Hartford, Richard Hess,

Gabrtelle and J. Donnay, Paul Sonn, Phelps

Meakor, Mary Lindenberg, Steve Feldman,

Eugene Boehne, John Woolston, Dave Slmen,

Harry Zaremba, Emmet Duffy, Matthew Fountain,Robert Holt, and Farrel Powsner.

N/D 3 A massless beam of length L is supported

by two stanchions at distances d1 and d2 from the

ends. The beam is loaded with point masses A, B,

arid C at the ends and midpoint. What Is thedownward force on each stanchion?

Robert Holt sent us a fine solution:

tF,

-i

The following solutions are from Darryl Hartman:Q—B3 K—B1

Q—KN3 K—K2

Q—Q6

or

— K—R2

B—B7 K—R3Q—R8

Also solved by Matthew Fountain, George Far-

nell, Ronald Raines, Robert Holt, Richard Hess,Randy Kimble. Winslow Hartford, and the proposer. Bob Kimble.

N/D 2 Given ABC is a right triangle; B is a rightangle; ABDE, BCFG, and ACHK are the squares

on the sides; and BQ Is parallel to AK. Determineif the three lines EC, BQ, and AF meet at a point Zas they appear to in the drawing.

To simplify the notation, say that the point masses

have weights A, B, and C. (If A is the mass, then

the weight is Ag, and A, B, and C must be re

placed by Ag, Bg, and Cg, respectively, In what

follows.) There are five forces acting on the beam;

three are the weights A. B, and C, and the other

two are the upward forces due to the stanchions.

Call these forces F, and F,. Apparently the beam

Is supposed to be at rest, so the torque about

.each stanchion is zero. The magnitude of the

torque about the left stanchion is

Ad, -B([U2)-dJ -C(L -d,) +F,(L -d, -d,).

Setting this equal to 0 yields

Ad, -B|U2] +Bd, -CL +Cd, +F,(L -d, -d.)= 0

(A +B +C)d, -(IB/2] +C)L +F,(L -d, -d.) =0

F, =[(V4B +C)L-(A +B +C)d,J/(L-d, -d ,).

Similarly, the magnitude of the torque about theright stanchion is

A(L -d.) +B([U2J -d.) -Cd, -F,(L -d, -d.)= 0,

and (his gives

F, = ((KtB +A)L -(A +B +C)d,J/(L -d, -d,).

Of course, by Newton's Third Law of Motion the

downward force on the left stanchion is((V4B +A)L -(A +B +C)d,y(L -d, -df).

and the downward force on the right stanchion isKViB +C)L -(A +B +C)d,y(L -d, -dt).

Also solved by Richard Hess, John Boynton. Irv

ing Hopkins, Howard Wagner, John Prussing,Haus Meier, George Piotrowski, James Reswlck,

John Woolston, Matthew Fountain, Robert Slater,

and Harry Zaremba.

The

Ben Holt

Co.^j^ d ^^

of FadWra lor On Enenjy

SpoctobSn m Gootfierraal

Ischnofagy

Ben Hal-37

afford A. PhSps. «2

201 Souih Late Avenue

PassdoM. CA 91101(213)684-2941

BrewerEngineering

Laboratories

Consul&ng EngineersExpcnmcntal Stress

AnalystsTheoretical Stress Analysis

Specialized Electro-Mechanxal Load Cdhand Systems. StructuralStrain Gago Ccndftnrungand Monitoring EquipmentRotating and StationaryTorquemotera

Given A. Brawcr '38Leon I. Weymouth '46

Manon. MA 02736

(617)7484103

CapitOl Coreufcng Civil Er^noen

Engineering BSfo**'41Corporation

17019

Gorham

InternationalInc.

Contract Recouch &DevelopmentMarket Arolyiss &Forocas&ng.

Commensal Development.tn

Chenscoh. Mmorab.Powder Metallurgy. Pulp 6Paper. Plartcri.

HUGH D OLMSTEAD.PhD -69P.O. B« 8Gorham. ME 04038(207)8922216Teiex 94-4479

Smoel956

TECHNOLOGY REVIEW A23

Syska &

Hennessy

Inc.

840 Memorial Dr.

Cambridge, MA

02139

TAD

Technical

Services

Corp.

Offices in:

Arizona

California

ColoradoConnecticut

Florida

Georgia

IllinoisKansas

LouisianaMaryland

Massachusetts

Michigan

Minnesota

Montreal

Debes

Corporation

Engineers

Mechanical/

Electrical/Sanitary

John F. Hennessy

■51

11 West 42nd St.New York. NY

10036

1111 19th St

NW

Washington. DC

20O36

575 Mission StSan Francisco. CA

94105

5901 Green Valley

CircleCulver City

Los Angeles, CA

90230 ""

Contract Technical

Services to Industry

and Government

for 26 years

Home Office:

639 Massachusetts

Avenue

Cambridge, MA

(617)868-1650

Missouri

Nebraska

New JerseyNew York

North Carolina

OhioPennsylvania

Quebec

TennesseeTexas

VirginiaWashington

Washington, D.C.

Wisconsin

Health Caro Consultant!

Design. Constnjcton.

Management

Sibaianes:

Charts N. Debes &Assoc. tncAkna Notcn Manor IncPark Sttathroocr

CorporationRocUcrd Convalescent

Confer Inc.Chambro Corporation

Charles N Debes 355668 Strathmore DnveRockiord. IL6U07

HID 4 Will the three hands of a clock (hours,

minutes, and seconds) ever divide the clock face

into three equal parts? If so. when?

James Reswick sent us an interesting analogy:

I'm not much on mathematical proofs, so I tried a

model clock in my Radio Shack PC2 computer.

The program runs the "clock" very fast but slows

down to calculate for each degree of second-hand

movement when the angle between the minute

and hour hands approaches 120°. At four times,

viz: 2:54:32, 5:05:27. 6:54:32 and 9:05:27, it

showed the second hand to be only 10° off when

the minute and hour hands were exactly 120°

apart. Thus there seems to be no solution to the

problem but ihere are four pretty close answers.

Reminds me of the old story about a mathemati

cian and an engineer. Each stood at one end of a

long room. At the other end stood a lovely young

lady. They were told to approach her by walking

half the distance, to stop, then to walk half the re

maining distance, stop, and to continue until they

reached the maid to receive a rewarding kiss. The

mathematician declared "But according to Zeno

I'll never reach her," and he left the room in dis

gust. The engineer started walking with alacrity

saying, "I learned about Zeno also—but I'll get

close enoughl"

J. Meier sent us a more conventional treatment:

The second hand advances 360° per minute; the

minute hand advances 6°per minute; and the hour

hand advances 0.5"per minute. The minute hand

gains 5.5° per minute 1= (11/2)° per minute] on

the hour hand. The second hand gains 354° per

minute on the minute hand. The first condition is

that the minute hand be I20°or 240° ahead of the

hour hand. Starting at X:00:00 (where the minute

and second hands are at 0° and the hour hand is

at X hours, or 30X°). it takes (SOX +120) - 2/11

and (30X + 240) - 2/11, or (60X + 240) /11 and

(60X + 480) /11 minutes, respectively, to meet the

first condition. This repeats every 12 hours, i.e., X

= 0. 1 11. The corresponding number of de

grees moved by the minute hand are (360 +

1440)711 =360(X +4)711 and 360(X +8)711,

respectively. As a check consider X = 7 and X =

3, respectively. The minute hand moves 360"after

7:00:00 and after 3:00:00 to come to 8:00:00 and

4:00:00, respectively, where it is 120° and 240°,

respectively, ahead of the hour hand. The second

condition is lhat the second hand be 120° or 240°

ahead of the minute hand. Starting at X:Y:00,

where the second hand is at 0° and the minute

hand at Y minutes, or 6Y° (X does not matter), it

takes (6Y + 120)/354 or (6Y + 240)/354 minutes,

respectively, to meet the second condition. Thisrepeats every hour, i.e., Y =0.1 59. The cor

responding number of degrees moved by the

minute hand are (6Y + 120)759 and (6Y +

240)759, respectively. As a check consider Y =

39 and Y + 19. In both cases it takes exactly one

minute to fulfill the second condition at 40 minutes

and 20 minutes, respectively, after the full hour.

For both conditions to be fulfilled simultaneously.

360(X + 4)/i 1 or 360(X + B)/11 must have a frac

tion in excess over a full integer equal to the frac

tion over a full integer of (6Y + 120)/59 or (6Y +

240)/59. respectively. But this is not possible for

the stated values of X and Y.

Also solved by Harry Zaremba, Richard Hess,

Robert Holt. Winslow Hartford, C. L. Baker.

Emmet Duffy, Matthew Fountain, John Woolston,

Dave Simen, Norman Wickstrand, Ken Haruta.

and the proposer. Alan Davis.

N/D 5 Six different numbers are selected arbitrar

ily from eight positive consecutive integers. The

resulting selection includes the smallest and

largest of the eight numbers and can be separated

into three pairs of numbers, each of which con

tains consecutive numbers. The sum of the six in

tegers is three times a number N, and the sum of

their cubes equals the cube of N. Find N and each

of the integers selected.

The following solution is from John Bucsela:

The eight consecutive integers may be written as

n, n +1 n + 7 for some integer n a 1. Thereare three ways to select six of these integers so

that the smallest and largest are included and so

that they may be divided into three pairs of con

secutive numbers. They aren, n +1,n +2, n +3|n +6. n +7n, n +1,n +3, n +4. n +6, n +7

n, n + 1. n + 4, n + 5. n +6. n+7

In the first case, the sum of the six numbers is 6n

+ 19. This is not divisible by 3, hence cannot

equal 3N. Likewise, in the third case, the sum of

the six numbers is 6n + 23, also not divisible by 3.

Thus only the second case can occur. Then, the

sum is 6n +21, so we conclude that 3N = 6n +

21, or N =2n + 7. Next.

n> + (n + 1)' + (n + 3)J + (n + 4)' + (n + 6)1 +

(n + 7)' = 6na + 63n' + 333n + 651.

Setting this equal to

NJ = (2n + 7)' = Bn1 + 84n* + 294n + 343, we

obtain 2nJ + 21 n' - 39n - 308 = 0

It is easy to see that the only possible positive in

tegral solution of this equation is 4. Thus N = 15

and the six selected Integers are 4,5, 7, 8,10, and

11.

Also solved by Richard Hess, Robert Slater,

Robert Holt, Howard Wagner, Steve Feldman,

Roger Milkman, Samuel Levitin, Howard Sard.

Richard Bernicker, Emmet Duffy, Matthew Foun

tain, David Simen, Frank Carbin. John Woolston,

Ronald Raines, Norman Wickstrand, and the pro

poser, Harry Zaremba.

Better Late Than Never

JUL 1 The answer supplied by the proposer was

correct, but I carelessly misread it.

A/S 4 Richard Hess has responded.

OCT 1 Richard Hess has responded.

OCT 3 Richard Hess has responded.

OCT 5 Richard Hess, Michael Jung, and Avi

Ornstein have responded.

Y1982 John Fine, Erik Anderson, Thomas Weiss.

Phelps Meaker, and Harry Hazard noticed that the

published solution was not optimal. A revised so

lution follows. To clarify the problem let me add

that we seek a minimum number of operations

(each addition, subtraction, multiplication, divi

sion, and exponentiation counts as one opera

tion); and among answers having equal numbers

of operations, we prefer those with 1, 9, 8, and 2

used in order.

1

2

3

4

5

6

7

8

9

10

12

13

14

15

16

18

19

20

21

22

25

27

28

30

33

35

]M

1"

1"

X 2

+ 2

18/9 X2

9 -■12+8

9/12 x8

18

1W

91

1"

12'

19

-9 -2

x8

-82

+ 9» - »>

-8 +2

1+9+8/2

191»

-8/2

x 8 x 2

18/2 +9

1 x 28 -r 9

1 --9+28

1 x 29 - 8

(19

19

18

1"

19

(119

-8) x 2

+ 8-2

x2 -9

x28

x2 -8

+ 2) X 8 + 9

+ 8x2

36 1

37 1

45 1

47V 1■49 1

50

54 (

57

64

70

8+9x2

x9 +28

8x2+9

9+28

X9B/2

+ 98/2

19 + 8) x 2

+ (9 - 2) x 8

• x8«

31-9-2

74 92 -18

75 91 - 8 x 2

81

82

84

85

87

88

89

90

91

92

95 <

98

100

18 X9/2

> x82

1 x 92 - 8

1 +92-8

x89 -2

1 +B9 -2

« x89

• +89

1 X 9 + 82

1+9+82

31 +8/2

l> X98

1 x9B +2

Proposers' Solutions lo Speed Problems

SD 1 There is one elegant and simple answer to

the problem that is quickly seen by those obser

vant people who do not overlook the obvious: 1.

They ask for the time tu required by the ball to fall

through the molasses; and 2. They write the

equation (called for in the problem). The total time

tT = tM + 2.73.

SD 2 Wrong.

A24 APRIL 1983

Syska &

HennessyInc.

840 Memorial Dr.Cambridge, MA02139

TAD

Technical

Services

Corp.

Offices in:

ArizonaCaliforniaColorado

ConnecticutFlorida

Georgia

Illinois

Kansas

Louisiana

MarylandMassachusettsMichigan

MinnesotaMontreal

Debes

Corporation

Engineers

Mechanical/

Electrical/Sanitary

John F. Hennessy'51

11 West 42nd St.

New York. N.Y.

10036

1111 19th St..

N.WWashington. D.C.

20036

575 Mission St.San Francisco, CA

94105

5901 Green Valley

CircleCulver City

Los Angeles, CA

90230

Contract TechnicalServices to Industry

and Governmentfor 26 years

Home Office:639 MassachusettsAvenue

Cambridge, MA

(617) 868-1650

Missouri

NebraskaNew JerseyNew YorkNorth CarolinaOhio

Pennsylvania

Quebec

Tennessee

TexasVirginiaWashington

Washington, D.C.

Wisconsin

Health Care Consultants

Dcsiqti, Constnirhon.

Management

SubsdiAncs.

Chafes N. Debes SAsxjc he

Abns Ncfeon Manor IncP<3rk StnithmocrCorporation

Rockford ConvalescentCenter Inc.

Chsrnbro Corporation

Charies N. Debes 355666 Strjthraore Dnvo

Rockloid. 0.61107

Puzzle Corner

Allan J. Gottlieb

This One Is

"Infinitely Hard11

Allan J. Gottlieb,'67, is

associate research

professor at the

Courant Institute of

Mathematical

Sciences of New fork

University; he studied

mathematics at M.I.T.

and Brandeis. Send

problems, solutions,

and comments to him

at the Courant Institute,

New York University.

251 Mercer St., New

York, N.Y. 10012.

As promised last issue, here's a descrip

tion of how we choose solutions for publi

cation.

As responses arrive during the month,

they are simply put together in neat piles,

with no regard to their date of arrival or

postmark. When it is time for me to write

the column, I first weed out erroneous

and illegible solutions. For difficult prob

lems, this may be enough; the most pub-

lishable solution becomes obvious. Usu

ally, however, many responses still re

main. I next try to select a solution that

supplies an appropriate amount of detail

and that includes a minimal number of

characters that are hard to set in type. A

particularly elegant solution is, of course,

preferred. I favor contributions from cor

respondents whose solutions have not

previously appeared, as well as solutions

that are neatly written or typed, since the

latter produce fewer typesetting errors.

Problems

M/J 1. Given the diagram at the top of

the next column, John Cronin wants

White to move and mate in four.

M/J 2. Frank Rubin sent us an "infinitely

hard" problem:

Differential equatbns crop up frequently,

usually first order, but sometimes second

order, rarely higher. For Review readers,

though, we go right to the limit: Find a

non-trivial, continuous, real-valued func

tion f(x) possessing continuous deriva

tives of all orders, and satisfying the

infinite-order differential equation

f =f +4f" +9P" +16f"" +

M/J 3. A pair of cryptarithmetic puzzles

from Thomas McNelly:

EVER

VWW

= .ONANDONANDON ...

.ONANDONANDON ...

IRIS

M/J 4. Larry Bell asks us about con

secutive primes:

Find three consecutive three-digit prime

numbers x, y, z, such that z - y = y - x

= 12. In other words, there are no prime

numbers between x and z, other than y.

Find four consecutive three-digit prime

numbers a, b, c, d, such that d - c = c -

b = b - a = 6. What is the largest string

of consecutive primes P,, P2,..., Pnsuch

that:

Pn - Pn - 1 = Pn - 1 - Pn - 2 = . . . =

P* " P, = 4.

M/J 5. Harry Zaremba notes that for any

triangle, the ratio of the sum of the

squares of the medians to the sum of the

squares of the sides equals a constant

ralbnal number. What is this number?

Speed Department

M/J SD1. We end with a problem from

Phelps Meaker:

To build a straight sidewalk 54'8" long,

there is a pile of 40 cast-concrete slabs,

formed as equilateral triangles. To

square off the ends, two extra 30°-60°

half-slabs are provided. What is the width

of the walk?

M/J SD2. A birthday quickie from John

Linderman:

On my birthday this year, my age was nor

the product of exactly two primes. How

ever, for a few previous birthdays in suc

cession, it had been. What is the largest

number of successive birthdays on which

your age can be the product of two

primes? Given that I just went past the

first such maximal sequence, how old am

I? If I live to be 100, how many such max

imal sequences will I celebrate?

A14 MAY/JUNE 1983

Edward R.MardenCorp.

Builders (or Industry.

Instrtubcns. Hotptlab.

Manufactunng Plants.Government and dovol-

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TECHNOLOGY REVIEW A13

Solutions

JAN 1. Unfortunately, the proposer has written

that his original analysis was in error and the prob

lem should be withdrawn. Richard Hess and DougVan Patter noticed that, in fact, only those de

clarers making the wrong play at trick S took 11

tricks.

JAN 2. On a map ol Massachusetts, each town is

connected to its nearest neighbor (assume no

ties). Show that no town is connected to morethan five others.

Daniel Seidman solved this problem by making

two key observations. First, since there were noties, any three cities form a scaline triangle (i.e.,all sides lengths unequal) and thus the longest

side would not appear on the map. Second, for a

. city A connected to six or more others there must

exist two neighbors B and C such that angle BAC

is less than or equal to 60°. Now it's easy. Triangle

BAC has angle sum 180° (or greater if on the

globe), so the largest angle cannot be BAC and

hence either line AB or AC does not appear.

Also solved by Avi Omstein, Leo Marten, NaomiMarkowrtz, and Richard Hess.

JAN 3. Drill a hole through a solid sphere (along

a diameter) whose radius is such that the height ofthe remaining ring is two inches. Find the volumeof the ring.

The following pair of solutions is from Avi Orn-stein:

Let the radius of the sphere be R. The height ofthe missing sections of the sphere (top and bot

tom) is R - 1. The cylindrical hole has a radius

which is one side of a right triangle, with 1" as the

other side and H as the hypotenuse.

The volume of the sphere is 4irrV/3.

The volume of the cylinder Is n-r'h, or w(R2 - 1 )2.

Each segment of the sphere is

wh'(3r - h)/3, or w(R - 1)f,2R + 1)/3.

The volume of the ring is therefore

4ffR'/3 - (2irR» - Zn) -2<2WRV3 -or 4ir/3 In3.

If the original sphere had a diameter of 2", the holewould be nonexistent, and the volume of the

sphere would give the same answer, as an

alternate way ol solving the problem. '

Also solved by Phelps Meaker, Ken Haruta,

John Prussing, Frank Carbin, Daniel Seidman,

Ronald Raines, James Reswick, Fredrick Hutch-inson, Richard Hess, Harry Zaremba, Naomi Mar-

kowitz, Norman and Ammi Spencer, Roger

Milkman, Norman Wlckstrand, Richard Marks,

Emmett Duffy, Leo Harten, David Evans and theproposer, Dean Edmonds.

JAN 4. In building a model, a man found that he

needed a 2:1 gear ratio. He had on hand only six

equal gears (like those shown below), yet he was

output

WB

able to obtain the desired ratio, using full tooth

conventional meshing. How?

A beautifully drawn solution shown at the bot

tom of the previous column was submitted by

Floyd Kosch.

Also solved by Richard Marks, Emmett Duffy,Chuck Coltharp, Richard Hess, Luigi Burzio, and

James Reswick.

Better Late Than Never

Y1982. Responses received from Robert Sack-he im and Harry Garber.

A/S 2. Richard Hess has found a solution if we

are not restricted to bishop moves.

OCT 2. Hillary Fisher and R. Morgan believe that,

for example, the sum TWELVE x*2 indicates thatthe value for TWELVE should be half the sum.

Their solutions are

13744

64948

64948

64948

208588

= 104294 x 2

84502

84502

68387

68387

68387

818387

1192552

= 298138 X 4

645

75054

75054

75054

75054

375915

= 125305x3

5394

5394

24947

24947

24947

24947

404947

404947

920470

= 184094x5

=W,

= - wB = - w,

= - wA + w,

= W, + W, = 2W,

OCT 3. Nancy Everds believes the cone contains

hot fudge and the sphere is ice cream. Thus melting must be considered.

OCT 4. Steve Chilton does not believe that a so

lution involving Bessel functions (or any other

series-defined functions) should be considered

closed form.

OCT 5. Jordan Wolk submitted an alternate solu

tion.

N/D 2. Harry Garber and Leo Harten have re

sponded.

N/D 3, N/D 4. Leo Harten has responded.

N/D 5. Harry Garber, Leo Harten, and William

Stein have responded.

1983 JAN SO2. Ken Fawcett believes that Van-

Patter's line of play is reasonable but not guaran

teed to succeed.

1983 F/M SD1. L. Steffens and Jordan Wolk

noticed that It is the lock that is 10 by 30 feet.

Proposer's Solutions to Speed Problems

SD 1. 27.733"

SO 2. Three in a row. If two of the years were

even, one would be divisible by 4. Since 4 is

flanked by 3 and 5, 4 itself cannot participate inthe sequence, and neither can 4x anything else.

The "middle number" in such a sequence must

therefore be 2 x p for some prime p. Browsing

around for such numbers flanked by the product of

two primes, one first hits paydirt at p =17, with

33 34 35

3x11 2x17 5x7

So I am (indeed) now 36 (a perfect square, some

one insisted on pointing out). Others (I hope Ididn't miss any) are:

p = 43 5 X 17 = 85 2 X 43 = 86 3 x 29 = 87

p = 47 3 X 31 = 93 2 x 47 = 94 5 X 19 = 95

So it looks like three by 100.

Sensory

Systems,

Inc.

Bart Johnson 75Sensory Systems,Inc.

104 Charles Street.Boston, MA 02114

617-354-0567Telex 940-536

Microprocessor-

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Design andDevelopment

From conception to

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Headquarters:45 RockefellerPlaza

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Boston Office:86 Coolidge Ave.Watertown, MA

02172

(617)926-5500

Alexander W.

Moffat, Jr.

TECHNOLOGY REVIEW A1J

TheBen Holt

Co.

Brewer

EngineeringLaboratoriesInc.

James

Goldstein&

Partners

ARCHITECTS

ENGINEERS

PLANNERS

S. James Goldstein '46Eliot W. Goldstein 77

225 Millbum AvenueMillbum. NI 07041(201)467-8840

Biochemistry

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Inc.

Engineers and Constructors

Phnningj and FeasibilityStudiesDesign and Constructionof Facilities lor the EnergyIndustries

Specialists In Geothorma)

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Consulting EngineerExperimental StressAnalysis

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FACILITIES:

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FOR HIGHTECHNOLOGY FIELDS:

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HUGH D OLMSTEAD.PhX '69

P.O. Box 8Gorham. ME. 04038(207)892-2216Telex 94-4479

Since 1956

James

Goldstein& Partners

S. James GoWsten>J6225 Millbum AverAteMilDburaNJ 07041(201)467-8840

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and Produrtscn FaeifctOG

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George A Homan. A.I.A.'65

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Newton, MA 02158(617)332-5427

Planning Consultants andDevelopers

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Sam Mintt. 79Anthony Mackall. 72

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500 Eighth Avenue.Suite 906

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Research, Developmentand EngineeringServices In the Electrical

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DECOTRAS.A.L.

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CcraJtnrj Erayneers

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A16 MAYfJUNE 1983

Puzzle Corner

Allan J. Gottlieb

How Many Ways

to Play Bridge?

Allan J. Gottlieb, '67,

is associate research

professor at the

Courant Institute of

MathematicalSciences ot New York

University; he studied

mathematics at M.I. T.

and Brandeis. Send

problems, solutions,

and comments to him

at the Courant

Institute, New York

University, 251

Mercer St., New York,

N.Y.. 10012.

Our son David had his first birthday a few

weeks ago (March 14), and he continues

to run all around our city apartment and

country house. But this activity (as well

as lots of others) has been somewhat

hindered by the heavy rains this April; I

conclude that New York will have a great

many flowers this May (and that Missis

sippi will be awash in blooms).

The published version of F/M 1 con

tained a serious misprint, and the (corrected) problem is reopened. Solutions tothe revised problem will appear in

November/December.

Problems

JUL 1 A bridge problem from Winslow

Hartford: Given the hands shown at thetop of the next column, South is to make

six spades, with the opening lead fromWest VQ.

JUL 2. Although this problem from JerryGrossman depends on the rules ofbridge, I am classifying it as combinatorial: The number of ways the play of a

hand of bridge can proceed obviouslydepends on the distribution and trumpchoice. Can some reasonably tight upper

♦ J 9 8 7 2

V5 4

♦ A5 4

+ AK2

♦ 654 *_

VQJ 10 9832 V 7 6

♦ 9 4QJ 10876♦ 9 6 f QJ 1087

♦ AKQ103

V AK

♦ K3 2

♦ 5 4 3

or lower bounds be found? How about an

average case analysis?

JUL 3. A pair of cryptarithmetic puzzles

from Avi Ornstein: Nine is a square, andwhile NINETEEN isn't actually a prime, at

least its smallest factor is greater than

150.

TWO + TWENTY = TWELVE + TEN

(The first three are all divisible by theirnamesakes.)

JUL 4. Anthony Stanton has a block of

wood with slots AB and CD, as shown. In

each slot is a wooden piece, P and Q,

that can slide in it. Also shown is a bar at

tached by loose screws to P and Q and

extended to a handle H. Is it possible to

rotate the handle through a full 360 degrees without having P and Q collide? If

this is possible, is the orbit of H an ellipse?

JUL 5. Our Japanese friend Nob

Yoshigahara sent us the following list of

the first 19 perfect squares containing

two distinct digits and not containing a

zero. What are the next two elements of

the sequence?

4x4 =16

5 x 5 = 25

6 x 6 = 36

7 x 7 = 49

8 x 8 = 64

IP3$

sins! !**■"*"

m

tasaa

tti

£15

See problem FIM 1, next page

12 x12

15 x15

21 x21

22 x22

26 x26

38 x38

88 x88

109 x109

173 x173

212 x212

235 x 235

264 x 264

= 144

= 225

= 441

= 484

= 676

= 1444

= 7744

= 11881

= 29929

= 44944

= 55225

= 69696

11

3 x 9 = 81

x 11 =121

CapitolEngineeringCorporation

Haley &Aldrich,

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Robert E. Smtfi '41

EuWQItl W OOQQ3 '56

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Cair£ndoe.MA02M2(617)492^460

TECHNOLOGY REVIEW A21

Gorham

International

Inc.

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Roman &AssociatesInc.

Planning

Innovations,

Inc.

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Contract RoGOarch &Development,Market Analyss &

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HUGH D OLMSTEAD.PhD €9PO Bon 8Gorhsm. ME 04038(207) 892-2216

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Since 1956

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George A Roman. AI A'65

Otw Gateway CenterNewton. MA 02158(617)332 5427

Planning Consultants and

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Tranrpcriabcn PlanningEnvironment Impact

fcccnomic and FinancialAnalysis

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Land-use PlanningMixed use Development

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Sam Minti. 79Anthony Mackal "72Joanne C Roche. 79

500 Eighth Avenue,Suilo 906

New York. NY. 10018

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Rcaoftrcri. DevelopmentAnd EnglnocrlnQ

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Spociaixn:Electric power systems.Electric transportation

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E. I. Stanberrj. 80T. vonHceenvinqo IV. 80

M.I.T. ALUMNI

CAREER SERVICES

GazetteA listing every two weeks

of jobs for alumni across

the country

We want more firms to

know they can list jobs in

the Gazette

We want more alumni mak

ing a job change to know

the Gazette can direct them

to job opportunities

Whether you have a job to

fill, or are looking for a job,

let us send you a copy of

the Gazette to show you

how It can help you

Call or write

Patricia O'Connor

Alumni Career Services

M.I.T., Room 12-170

Cambridge, Mass. 02139

Tel: (617) 253-4737

F/M 1. Elliot Roberts and Richard hfess

noticed that no solution was possible. In

specting the proposers' solution, I am

able to deduce that the Black queen and

pawns should be White. The problem is

therefore reopened; White is to mate in

three.

Speed Department

JUL SD1. Smith Turner writes: I am told

that a clock (which I am not shown) is 1

hour and 20 minutes fast, that the minute

and hour hands are interchanged, and

that it is placed on the wall behind me. On

the wall in front of me is a mirror I can't

see, but I see the image on this mirror re

flected from a plate glass cover on my

desk top. The desk top image is shown

below. What is the correct time?

JUL SD2. You are given a three-pint jug

and two five-pint jugs, all three empty,

and an eight-pint jug full of wine. Divide

the wine into two equal portions.

Solutions

F/M 2. Let T be the torus obtained by rotating the

circle (x - 2) + y = 1 in the x-y plane about the

y-axis. Let B be a plane tangent to the torus T at

the point (0,0,4). Find the volume and surface

area of the small region obtained by slicing T with

P.

The following solution Is from Matthew Foun

tain: The smaller part of the torus has a volume

approximately 13.03 and a total surface approxi

mately 33,029. The flat portion of the surface is

exactly 8.

A22 JULY 1983

Ti

l

The volume ol a figure of revolution is JO wr dr,where w is the width parallel to the axis of the fig

ure at distance r from the axis, and 0 is the angle

of revolution of that width. The width of the torus isw = 2V1 - (r - 2)» and 9 = 2 sac-'r, making the,smaller part of the torus contain a volume equal to

4f* rVI - (r - 2)' sec-'r dr = 13.03

when evaluated by Simpson's rule. The flatsurface is

S, = / w dx, where

x = Vr1 - 1 and dx = r/[Vr" - 1) dr. Then

S, = 4|>rV1 - (r - 2jV(VF"^1] dr

= 4|»rv9^7/vm dr.

Substituting t« = r + 1 results in

S, = 8f* (t» - Dv^Tdt =2

| -t(4 -«•)"• lf =8.

The curved surface Is

Sc = I r 6 ds, where

(ds)' = (dw/2)« + (dr)«

= [(r - 2)'(<Jr)']/[1 - (r - 2)'j + (dr)'

= (dr)«/l1 - (r - 2)'\. Then

Sc = 4/j (r sec-'r|/[V1 - (r - 2)»|dr.

The substitution of r = 2 + sin <b converts this to

Se =4f I (s + sin <*)sec-'(2 + sin <4)d« =

25.029 when evaluated by Simpson's rule.

F/M 3. Solve the simultaneous equations:x«y = (y» + 1 )x'

y*x =9(x» + IJy3

The following solution is from Charles Sutton:

If we discount the trivial solution x = 0, y = 0, we

may divide the equations by x1 and y3, respectively, to obtain

x'{xy)=y« + 1

<1'

<2>

y)( )

Now make the change of variables

fix = v

from which we obtain

y' = uv <3'and equations (2) become u'/v = uv + t and u'v =

9(u + v)/v which, when cleared of fractions, giveu2 = uv* + v (4)

u'v' = 9u + 9v (5)

Multiplying the first equation by 9 and subtracting

the second gives 9u* - u'v' = 9uv» - 9u which,when solved for v», gives

v. _ 9u'

From (4) we have v = u' - uv' = u» - (9u' +

9u)/(u + 9) = (u> - 9u)/(u + 9) so V = (u* - 18u«

+ 81u')/(u + 9)' and substituting this in (6) gives

(u« - 18u« + 81u')/(u + 9)' = (9u + 9)/(u + 9)which when cleared ol fractions gives u« - 18u« +81 u' =9u' + 90u + 81, or

u« - 18u' + 72u' - 80u -61 =0 (8)

We see from (1) and (2) that xy, and hence u,

must be positive. Equation (7) is found to have

one positive root, located between 3 and 4, and

using a calculator, the value of this root is found to

be u = 3.888908631. It follows from (3) that v

must be positive and from (6) we find v =

1.847647602. Knowing u and v, we can calculate

x» and y' from (3). Finally, we see from (2) that x

and y must have the same sign. Thus the nontriv-

ial solutions are

x = -1.45078915

y = -2.68054709

Relno Hakala wondered what MACSYMA (a

noted computer software system for algebraic

manipulation) would do on this problem. As it

happens. Leo Harten did use MACSYMA on hissolution, and he reports thai after 45 CPU sec

onds, using Its standard solver, MACSYMA found

the three solutions given above and ten involving

complex numbers. The proposer Norman

Wlckslrand notes that for a very similar (in appearance) set of equations

(X« + 1)Y =(Y» + 1)X>

the solution is "solvable by radicals." Readers

desiring a copy of Mr. Wickstrand's solution to thisshould write to the editor.

Also solved by Harry Zaremba, Emmet Duffy,

Richard Hess, Matthew Fountain, Irving Hopkins,Frank Carbin, and Jordan Wouk.

F/M 4. A pilot flies south over a spherical earth a

distance D, flies due east a distance D, and thenflies due north a distance 0, arriving back at pre

cisely the starting point. For D equal to the radius

of the earth, find all solutions for the startinglatitude.

The following countably infinite number of solutions are from Avi Ornstein:

The most obvious solution would be to start at theNorth Pole. Flying D miles south, then due east,

and then north would return to the North Pole. An

infinite set of solutions are also possible, however. They would entail starting at a point, flying D

miles south, then circling the world an integralnumber of times In 0 miles (thereby being at the

same point), and then returning 0 miles north

again to the starting point. The formula for the

circumference of a circle drawn on the surface ofthe earth is 2 D sin (r/D), where D is the radius of

the earth and r is the radius of the circle. Center

ing the circle at the South Pole, we can rearrange

the equation to be r = D arcsin (1/2 n). The set of

solutions would then be D + r, where n is set for

the number of times the pilot would circle the

world. The fact that the world is not actually asphere means that slight modification would be

required. Likewise, the length of a degree of

latitude varies between the equator and the poles,

but the error would not be too significant.The first seven distances north of the South Polehappen to be 1.160D, 1.080D, 1.053D, 1.040D,

1.032D, 1.027D, and 1.023D. The answer is ap-

x = 1.45078915

y = 2.68054709and

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proximalely (1 + 0.160/n)D, which should be'accurate enough. >

Also solved by David Evans, Phelps Meaker,

Jordan Wouk, John Woolston, Harry Zaremba,

Richard Hess. Matthew Fountain, Norman

Wtekslrand and Emmet Duffy.

F/M S. A vertical plumb line is hung down the cen

ter of a one-mile-deep mine shaft in Borneo. A

rock is dropped from next to the top of the plumb

line. How far from the bottom of the plumb line will

the rock land, if we ignore friction?

Our final solution is from John Prussing: Con

sider a coordinate system with its origin at the

center ol the earth. Because Borneo lies on the

equator, let the x axis point out the mine shaft, the

z axis point out the North Pole, and the y axis

complete the right-handed frame. This frame ro

tates with the earth, and to someone looking down

the shaft, the y axis points eastward. The absolute

accoloralion of the rock is -ge,, where g is the

(constant) gravitational acceleration and e, is a

unit vector along the x axis. The acceleration vec

tor of the rock relative to the rotating frame a is

given by

a, = -ge, - wx(wxr) - 2xv,

where w is the angular velocity vector of the earth,

r is the position vector of the rock from the center

of the earth, vr is the velocity vector of the rock

relative to the rotating frame, and x represents the

cross product. From Ihis equation the components

of a, are found to be

d'x/dt' = w'x + 2w dy/dt - g

d'y/dt' = w'y - 2w dx/dt

d'z/dt' = 0

where w is the magnitude of the angular velocity

of the earth and x, y, z are the components of the

position r. The solution to these equations for ini

tial conditions x(0) = earth radius = R, y(0) = z(0)

= 0, and all initial components of velocity relative

to the rotating frame = 0 is

x(t) = (R - g/w')(cos a + a sin a) + g/w1

yjt) = (R - g/w')(a cos a - sin a)z(t) a o

where a = wt. The boundary condition at the final

time T is x(t) = R - 1 miles. Using a value of earth

radius R = 3960 miles, the value of a which satis

fies the boundary condition is 1.320 x 10~' and

y(t) = 4.647 feet. Thus, the rock lands this dis

tance east of the plumb line.

Also solved by Sidney Shapiro, David Evans,

Ronald Raines, Bob Pease, Harry Zaremba,

Emmet Duffy, Norman Wickstrand, Matthew

Fountain. Richard Hess, John Woolston and the

proposer, Bruce Calder.

Better Late Than Never

19B2 OCTS. David Oreyfuss found a simpler so

lution.

N/D 3. Sidney Shapiro has responded.

N/D 5. Sidney Shapiro has responded.

19B3 JAN3. Rick Decker has responded.

Proposer's Solution to Speed Problem

JUL SD2. The following diagram demonstrates

the process:

uInitial

8-5

5-3

3-5

8-5

5-3

5-5

3-8

0

0

3

0

0

3

3

0

0

0

0

3

5

2

0

0

0

5

2

2

2

2

4

4

8

3

3

3

1

1

1

4

A24 .JULY 1983

Puzzle Corner/Allan J. Gottlieb

Now Math Is for Money

Allan J. Gottlieb, '67, isassociate research

professor at the CourantInstitute of

Mathematical Sciencesof New York University;he studied mathematicsat M.I.T. and Brandeis.Send problems,solutions, andcomments to him .it theCourant Institute, NewYork University, 251Mercer St., New York,N.Y., 10012.

A recent speed problem has provoked

Albert Mullin to offer cash prizes for so

lutions to two related problems. Such

individually funded prizes are not un

precedented; indeed, the famous

mathematician Paul Erdos has been

doing it for years. Mr. Mullin writes:

"John Linderman's speed problem

M/J SD2 (Mayl)une, 1983, p. AM) has

other interesting aspects that may

amuse the readers of 'Puzzle Corner.'

Not only are products of two distinct

primes connected in nontrivial ways

with M.I.T.'s version of public-key

cryptosystems (so-called RSA public-

key cryptosystems), but the presentyear and the past two years in the ordi

nary calendar (yet another three con

secutive numbers) are products of two

distinct primes, too. Thus:

1981 = 7 x 283

1982 = 2 x 991

1983 = 3 x 661

Another such triple of years occurred

during World War II:

1941 = 3 x 647

1942 = 2 x 971

1943 = 29 x 67

Surely such triples of consecutive num

bers have a role in a deep study of the

Cabala!

"I will donate $50 to the M.I.T.

Alumni Fund for a proof that there exist

infinitely many triples of three consecu

tive positive integers which are prod

ucts of two distinct primes.

"Further, it can be shown that if a

positive integer n is a product of two

primes that differ by d a 1, then

^(n)fr(n) = (n - d - l)(n + d - 1),

where <p is Euler's totient and <r is the

sum-of-divisors function. I will donate

$100 to the M.I.T. Alumni Fund for a

proof that the converse holds for infi

nitely many d s 1."

Problems

A/S 1 We begin with a chess problem

from Bob Kimble:

pawn, perhaps he should win—not just

draw. But watch out for the Black central pawn mass.

A/S 2 Donald Richardson asks us a var

iation on a well known mathematical

game, the "2-3-4-5 coin game." He

writes:

To play the game, 14 coins are ar

ranged in four horizontal rows with 2,3,

4, and 5 coins, respectively:

Two opponents, "A" and "B," take

turns picking up any one or more coins

from any one horizontal row until one

opponent wins by leaving the last coin

for the other opponent to pick up. If

"A" starts, there will be no way for "A"

to win regardless of his first move, un

less "B" fails to make the right moves

thereafter. The proble'm is to identify

how few and what configurations "B"

can leave for "A" on "B'"s first move

(after any starting move by "A") so that

"B" can win, regardless of any sub-

'Sequent move by "A."

A/S 3 The following puzzle made its

Technology Review debut in 1942 as part

of an advertisement for Caldron Prod

ucts, Inc.:

The diagram indicates schematically a

little problem encountered in our shop a

short time ago.

h- x

•-B *\ •

White is to play and draw. At first it

looks too easy: with material balanced

and White possessing the only passed

A member A moves on rollers, without

slipping, from the solid-line position to

that shown in dotted lines. What is the

value of X in terms of the lengths A and

B?

A/S 4 A geometry problem from Mary

Lindenberg:

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In the drawing, C is a point on the

semicircle with AB as diameter. MN is

tangent to the semicircle at C. AM and

BN are perpendicular to MN, and CD is

perpendicular to AB. Show that CD =

CM = CN and that CD2 = (AM)(BN).

A/S 5 A numeric problem from Susan

Henrichs:

Which integers X have the property that

9X is the same as X with the digits in re

verse order? There is one other integer

multiplier (besides 9 and trivially 1) that

reverses digits for an infinite number of

integers. What is this multiplier and

what are the multiplicands?

Speed Department

A/S SDl Art Delagrange proposes the

following variation on FM SDl:

If a 10-by-30-by-l-foot flat-bottomed

rectangular barge is in a lock and a 100-

cubic-foot chest with a specific gravity

of 4 is thrown overboard, what is the

change in water level?

A/S SD2 We close with a bridge quickie

from Doug Van Patter:

With the hands shown, you and your

partner have reached a six-no-trump

contract:

7 32

QJ 98

V A

♦ A

*QJ 5VK 5 4

♦ 10 6 2

♦ A 9 64

The bidding: one diamond, two clubs

(you), four no-trump, five diamonds

(one ace), six no-trump. The <fc8 (open

ing lead) is taken with your 4&Q- You

play the ^10—East shows out! You take

a second finesse and cash the +K—

West shows out! Back to your hand with

the VK. You cash the +A, discarding a

heart from dummy. Now you finesse

diamonds for the third time. What is the

best shot at making this contract?

Solutions

APR 1 Find a bridge deal such that four spades,

when played from any position, will be set at least

seven tricks (i.e., the defense will make four

spades).

L. Steffens liked this one because it was a prob

lem that he could handle but the expert among his

bridge buddies couldn't. The following solution is

from Lawrence C. Kells:

I was fascinated to see that this is a retrograde

A22 AUCUSTVSEPTF.MBHK 1983

bridge problem—that is, instead of presenting a

hand and asking how to reach a particular result,

you give the result and ask how to reconstruct the

hand. 1 have doodled some on this kind of problem

so I immediately tackled this one. I guessed that the

key was that one player had the top spades and

high cards in other suits to cash after drawing

trumps if he were on lead, but that he would lose

his high cards to a crossruff if he weren't. The solu

tion then fell into place:

♦ J 10 9

♦ J 10 9 8 7 6

♦ AKQ|

V2

♦ 5

♦ 5

4 3 2

4 3 2

4 3 2

♦ 87 6

V J 10 9876543

♦ -♦ 10

♦ AKQ

VAKQ

♦ AKQ

49876

If North or South is declarer, a dub lead allows

West to take four tricks, after which the defense

cross-ruffs six tricks in diamonds and hearts, while

South follows and North cannot overruff. South

gets only his three top spades. If East or West is

declarer, a spade is led and South draws trump in

three rounds. Then South can cash his six side-suit

winners. North's remaining trump is the tenth trick

for the defense.

Thus four spades from any position is defeated

seven tricks.

L. Steffens and Stephen Canter noticed that

slightly rearranging the diamonds (give North the

♦Q and «>J, West the *3 and South the *2) ena

bles North-South to set the contract one more trick.

Also solved by Matthew Fountain, Phillip

Dangel, Doug Van Patter, and the proposer, Arthur

Polansky.

APR 2 Shown below is a simple word square:

ACE

PAR

ERA

We could fill the plane with such word squares so

that every cell of an infinite grid lay at the intersec

tion of two words. However, such an arrangement

would have two defects: (1) the pattern would be

repetitive, and (2) the words would not all be inter

connected. Show how these defects can be fixed;

i.e., fill the plane with an infinite number of words

so that the pattern does not repeat in any row or

column, every cell lies at the intersection of two

words; and every word is connected to every other

one by a chain of intersecting words. Do not use

any two-letter words.

Matthew Fountain gave a solution involving the

group-theoretic definition, and Richard Hess sub

mitted the following:

n k

ATE...ATEASEATE...ATEASE

TE...ATE

E...ATE

ATE

TE

A

The above pattern repeated indefinitely with n and

k varied randomly (between 1 and 10, say) will

satisfy the conditions of the problem:

T: Always at the crossing of ATE and ATE.

A: At the crossing of EAT and EAT or TEASE and

TEASE.

S: At the crossing of TEASE and TEASE.

E: At the crossing of TEA and TEA or TEASE and

TEASE.

APR 3 One afternoon in a swamp Sneaky the

snake, whose mass is M, and length (,. was resting

on the left end of a log of length L and mass ML

when he suddenly remembered that he was hun

gry. After a few moments, he found Freddy the

frog sitting on a large disk-shaped leaf of radius R

and mass Mo h cm. away from the center of the

disk. Initially the disk was 4/5 (, cm. away from the

right end of the log. Sneaky started moving toward

Freddy, and Freddy (who was ignorant of the laws

of physics) started jumping for his life. However,

when Freddy remembered that Sneaky had his best

friend for breakfast that morning he was overcome

by rage and desire for revenge, and he turned

around to charge against Sneaky. Normally it

would have been a dumb move, but the laws of

physics reward the courageous: when Freddy stop

ped at a point on the disk and realized that his at

tempt was futile if not suicidal, the disk was just out

of the reach of Sneaky, who was on the right end of

the log by that time. Sneaky could not swim, so the

good guy was saved and the bad guy had a hungry

afternoon. Can you locate the spot where Freddy

stopped? Assume all motions occurred on a line

and the water offered no resistance. Both the log

and disk have a uniform mass density.

The following solution is from Harry Zaremba:

Ms

t

L

y

MF

L8 T

Vih|i

+ M0

In the drawing, assume Mr is the mass of the frog

and let x, y, and z be the distances moved by the

frog, log, and leaf, respectively, from their original

positions. If the velocity of the snake along the log

is V, = lit, the impulse forces between the snake

and log will impart a velocity of VL = y/t to the

combined mass of the snake and log. The distance y

is in the opposite direction to the snake's motion.

By conservation of momentum,

M.V. = (M. + MJV,

Substituting the velocities,

M.L = (M, + M,Jy, or

y = M.U(M, + ML).

Applying the same principle to the frog-leaf sys

tem,

Mfx = (M, + Mo)z,

in which z is the distance moved by the combined

mass of the frog and leaf in a direction opposite to

x. Thus,

z = M,xl(M, + M,,).

From the figure,

y + z + 4L./5 = L,.

Substituting y and z, the solution for x becomes -

x = (M, + M,)/M, • (L./5 - M,U|M, + MJJ.

Also solved by Leo Harten, Harry Zaremba,

Michael Jung, and the proposer, Ming Chung.

APR 4 Assuming the sudden onset of steady rain

fall, will a person remain dryer by walking or run

ning any given distance to shelter?

Richard Hess believes in running in the rain. He

writes:

If you can orient your body at will it is best to run as

fast as possible.

Model yourself as a long rectangular solid with end

cross area A. If you can move at velocity v you

should orient yourself at

fl=tan"'v,,

so as to only get the small end wet. The total water

received is that in a volume

V = Atv/sin ft

where t is the time spent in the rain. But

t = d/v

where d is the traveled distance; and

V = Ad/sin ft

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This becomes less as 8 gets larger, and thus

v = v „[„ tan 8 »

should be made as large as possible.

Matthew Fountain notes that with an umbrella it

is best to go slow and notes that when he is stuck in

the rain it's the inside surface of his glasses that gets

wet. Although Fountain says his friends dispute

this, I have the same experience. Judith Longyear

observes that "no animal moves slowly through

rain!"

Also solved by Harry Zaremba.

APR S Find positive integers a, b, and c such that

a' + b' = cs.

Although no one was able to prove that all so

lutions had been found, David and Nancy Leblanq

(among others) found infinitely many. They write

what they call "a computer scientist's solution":

Assume a, b, and c are powers of 2. This transforms

the problem into

2" + 2" = 2".

(This limits the solution space.) 2n must equal 2"

in order to sum to a power of 2 (in this case 2").

Thus 3x = 4y.

2" + 2" = 2" + 2" = 2"*1 = 2"

Thus 3x + 1 = 5z. 3x = 4y for all q such that x = 4q

(and then y = 3q). 3x + 1 = 12q + 1 = 5z, which has

integer solutions for q = 2, 7, 12, 17, ... This pro

vides a set of solutions for aJ + b' = c\

For q = 2, x = 4q = 8, y = 3q = 6, z = —" = 5

a = 2« = 256, b = 2' = 64, c = 2' = 32.

Also solved by Ronald Raines, (udith Longyear,

Richard Hess, Harry Zaremba, Leo Harten, Irving

Hopkins, Ruben Cohen, Anthony Beris, Sidney

Shapiro, Dick Boyd, Charles Sutton, David Evans,

and Robert Slater.

Better Late Than Never

NS 10 B. Laporte has some further partial results.

1981 OCT 4 Matthew Fountain found an exact so

lution:

V5" 4 5 1/V6\ ,/7\

This agrees to seven digits with Harry Zaremba's

approximate solution. Copies of Mr. Fountain's

derivation can be obtained from the editor.

1982 OCT 2. Matthew Fountain and an anonymous

computer found the missing solution:

E

E

S

S

S

E L

T W

I

I

E

E

E

E

E

G

C

V

V

V

V

L

H

H

E

E

E

E

V

T

T

N

N

N

N

E X4

S

2

8

8

6

6

6

1

9

4

4

8

8

8

8

8

5

5

3

3

3

3

1

0

0

8

8

8

8

3

2

2

7

7

7

7

8X4

Proposers' Solutions to Speed Problems

SD1 None, as the barge had been overloaded by 33

percent and would already be on the bottom.

SD2 Since East has eight dubs, there's a good

chance of only two hearts in the East hand. Cash

the 4>A and 4>K; if East follows he can have only

two hearts originally. Cash the VA and throw West

in with the last heart, for an end-play in diamonds.

If East has only two spades, hope that West has the

high heart honor after the VA. This was hand 28

from a tournament at Cherry Hill, N.J., on August

21, 1982. The end position was:

Dummy:

*AK

VA7

♦ AQ

West East:

*7 4 49 6

V Q 10 V J

♦ K 7 ♦ Q J 10

After the *A and 4tK and VA, West is end-played

in hearts. He has to lead a diamond back at trick 12.

Some declarers found this line of play; unfortu

nately, my partner wasn't one of them!

AM AUGUST/SEPTEMBER 1983

Puzzle Comer/Allan J. Gottlieb

Allan J. Gottlieb, '67, isassociate research

professor at the Courantinstitute of

Mathematical Sciencesof New York University;he studied mathematicsat M.I.T. and Brandeis.Send problems,solutions, andcomments to him at the

Courant Institute, NewYork University, 251Mercer St., New York,N.Y. 10012.

Fannie Dooley

Will

Gefozzle You

Since this is the first issue of a new

academic year, we once again review

the ground rules for "Puzzle Corner":

In each issue we present five regular

problems (the first of which is chess- or

bridge-related) and two "speed" prob

lems. Readers are invited to submit so

lutions to the regular problems, and

three issues later one submitted solutionis printed for each problem; we also list

other readers whose solutions were successful. In particular, solutions to the

problems you see below will appear in

the February/March issue. Since I must

submit that column sometime in

November (today is July 20), you should

send your solutions, to me during the

next few weeks. Late solutions, as well

as comments on published solutions,

are acknowledged in the section "Better

Late Than Never" in subsequent issues.

For "speed" problems the procedure

is quite different. Often whimsical,

these problems should not be taken too

seriously. If the proposer submits a so

lution with the problem, that solution

appears at the end of the same column

in which the problem is published. For

example, solutions to this issue's

"speed" problems are given below.

Only rarely are comments on "speed"

problems published or acknowledged.

There is also an annual problem,

published in the first issue of each new

year; and sometimes we go back into

history to republish problems whichremained unsolved after their first ap

pearance.

All problems come from readers, and

all readers are invited to submit their favorites. I'll report on the size of the

backlog, and on the criteria used in

selecting preWems-for publication, in afuture issue. \

Problems

OCT 1. We begin with a seven-card

bridge problem from Emmet Duffy.

South is on lead and is to take all seven

tricks against the best defense:

¥2

♦ A 4 3 2

65 8 2

♦ 10

♦ 83

* A

VK

♦ 5

A A

10 7

10

♦ Q

♦ QJ

OCT 2. The next (presumably serious)

offering is from Robert Pease:

But of course, Fannie Dooley loves toBaffle people. And Gefozzle them, too.

You see, Fannie loves vanilla, but not

chocolate. She loves Allan J. Gottlieb

but not Robert Pease. She likes doors

but not windows. She likes Mississippi,but not Alabama, California, or New

York. . . . Also, she loves Arrowroot

cookies . . .but doesn't care for tapioca.I told a couple of my friends, Lenna

Mooree and'William Llewellyn, and theyfigured it out. Fannie also loves the

Massachusetts Institute of Technology

and Harvard College, but not Princeton

University . . . she likes cherry-wood

furniture but not mahogany. Now do

you see why? It's a fun game for people

of 5th-grade mentality ... Of course,

you have figured this out, but if you

can't figure out why Fannie Dooley pre

fers all this hoop-la, to plain ordinaryBLAH, ask a VOODOO doc'tor. '

OCT 3. Benjamin Madero proposes a

variation on 1982 N/D 3; by writing "I

think a more interesting problem would

result if the loads and reactions on the

beam were inverted, that is, two con

centrated loads and three reactions."

Then the problem wouldi»e:

A massless beam of length L is sup

ported by three stanchions A, B, and C

at the ends and midpoint. The beam is

loaded with point loads F, and F2 at dis

tances d, and d_. from the ends. What is

the downward force on each stanchion?

-L/2- -1/2-

OCT 4. Greg Schaffer wants you to

show that for all N 2 3

= 0

OCT 5. John Woolson proposes six root

extraction problems (the last of which

he claims is not for the faint of heart). Inall the problems each x is to be replaced

with a digit (duplicates permitted), thenumbers are base 10, no leading zerosare allowed, and no zeros are allowed inthe roots themselves.

xxx

''v'xxx xxx xxx

xxx

X XXX

X XXX

xx

xx

Vxx

XX

Vxx

XX

X

XX

XX

XX

X

X

XX

XX

XX

X

X

XX

XX

XX

XX

XX

XX

XX

XX

XX

XX

XX

XX

XX

XX

xxx xxx

xxx xxx

X XXX

■\Kxxx xxx xxx xxxxxx

X XXX

X XXX

xxx xxx

xxx xxx

XX XXX XXX

XX XXX XXX

xxxx

'v'xxxx xxxx xxxx xxxx

xxxx

XX XXXX

XX XXXX

XX XXXX XXXX

x xxxx xxxx

xxxx xxxx xxxx

xxxx xxxx xxxx

Speed Department

OCT SD1. Smith D. Turner (also

known as J dt) asks what the following

three numbers have in common:

1. 371 288 574 2 ...

237; 581 208 759 3 ...

3550. 260 181 586 5 ...

OCT SD2. C. Baker has 5-, 11-, and

13-pint jugs, all empty, and a full 24-

pint jug. How does he divide the liquid

into three equal portions?

A22 OCTOBER 198J

Alan O. Ramo

Michael). SchafferJotH E. SchindallCharles C. SchumacherPeter T. Van AkenGrant M. Wilson

1964

F. Michael ArmstrongWayne F. BTleUsLeslie M. Boring, Jr.Richard A. CarpenterLeonard Chess

Ernest M. CohenRonald H. CordoverJohn P. OownieMichael B. Godfrey{ohn N. Hanson

.ester L. Hendrickson

Roger L. Hybels

Mark JosephLeon M. KaatzBrian R. Kashiwagi{oseph F. Kasper, Jr.

oseph L. KirkGlenn A. LarsonDonald S. LevyStephen B. MillerAlton B. Otis, Jr.Peter J. SherwoodJay M. TenenbaumThomas C. Vicary

1965

Arnold R. AtomsWilliam R. BrodyEdward A. BucherArthur A. BushkinW. David Carrier, IIIL. Scot Duncan

Howard M. Ellis *Peter C. CersrbergerJohn J. Golden, Jr.George L. HadleyDawn Friedell JacobsWilliam N. Kavesh

Louis A. KleimanPeter A. KlockAlan C. LeslieSteven B. LipnerJohn A. Ottesen

Robert B. ReicheltJohn D. C. RoachEmile SabgaCregory L SchafferDonald L. ShulmanRobert L. Silverstein

Douglas C. SprengRichard L. St Peters

G. Wayne ThurmanCarol E. Van AkenMichael G. WeissRonald WilenskyStephen L. WilliamsDavid L. Yuille

19*6

Michael R. AdlerArthur N. BoyarsRichard R. BradyPaul A. BranstadWilliam L. BunceRichard Y. ChungRichard T. CockeriUPeter M. Cukor

Ralph M. DavisonSteven H. Disman

Logan L. DonnelBert E. Forbes

Victor K. FungPhilip M. Jacobs

Michael D. KinkeadRobert D. LargeGerald F. MadcaHarry C. MoserHenry H. Perritt, Jr.Ralph G. SchmittJoseph I. SmullinJoseph W. SulUvan. Jr.Frank E. Surma, Jr.

William R. Tippell. II

John Torode

1967

John AcevedoDonald A. Belfer

James W. OrterWilliam L. Caton. Ill

W. Thomas ComptonJohn M. Davis

David A. DillingCarl B. DoughtyAlan B. Hayes

Lutz P. A. HenckelsEdson C. HendrtcksArthur C. KwokWilliam E. Murray, Jr.John S. PodulskyJohn C. H. ReykjalinChet SandbergJohn M. ShureltJoel M. Steinberg

Arthur S. Warshaw

1968

Harvey AllenPlatte T. Amsrutz, IIIPaul F. Bente, 111William E. Orison

Samuel A. CohenClaude L. GerstlePaul A. CluckDaniel M. GreenPeter GrootJohn P. Linderman

Scott P. Marks, Jr.Juan M. MeyerCharles B. Miller. Jr.

Kenneth P. MorseWilliam M. ParksJoel P. RobinsonLeonard H. SchrankJonathan D. ShaneMichael Weinreich

1969

Dariush AshrafiMelvyn P. BasanWilliam P. BengenRobert A. Bemtson

Denis A. BovinMarc DavbBruce R. DonathPaul D. EvansMatthew M. Franckiewicz

Anthony GeorgeAndrew C. GoldsteinBruce L. HeflingerKenneth R. HornerJoseph A. HortonBernard E. KleinAaron Kleiner

Alan K. KudlerCarl W. Kuhnen. Jr.

Michael W. LairdSamuel E. PolancoR. Frank Quick, Jr.

Franklin P. RogersChristopher R. RyanMichael Sporer

Michael P Timkolames P. Truitt. Jr.Hal R. Varian

Hing Y. Walt

1970

Gregory K. ArensonKaren ArensonDavid S. Bann

Wendell C. BraseGerald L. BrodskyJames C. BronfenbrenncrJames L. CaldwdlDaniel R. CherryEric K. demonsStephen F. CooperHarry D. FeldmanLinda L. FurrowRobert F. GonsettJohn C. Head IIIC. Gordon HunterRichard W. lhrieWilliam C MichelsWilliam B ParsonsDavid T. PattenAnthony C. PicardiChristopher L. Reedy

James B Rolhnie, Jr.

Stephen R. TakeuchiWaller Yorsz

1971

Richard A. Aparo

R. A. Castro AlpizarThomas C. KellyRobert D. Marshall, Jr.

Robert N. SchulteLaurence StorchCharles W. Werner

Philip R. Widing

1972

Bradley C. BilletdeauxJohn M. BisseUjack E CaterLeonidas P. ColakisDavid A. Davis

Marshall 8. GoldmanG. Paul Hendrickson, Jr.Donald H. Layton

Hampton PittsJames W. RoxloGary J. SchuitemaHikaru P. ShimuraJohn W. Taylor

Robert E. Zahler

1973

John R. Bertschy

John R. GershDebra JudelsonSamer S. KhanachetPatrick A. MarcotteStephen P. MiUerC. Timothy RyanPhilip M. SadlerJ. Alexander StevensCynlhia Day StrattonJohn W. Vander Meer, 111

1974

James Richard AndrewCharles Shaded BrunoIan FisherRoderick John Holland, Jr.

Niels Eric MortensenJohn Emery PlumGary David RaymondFrank M. SaukJay W. Van Dwingelen

1975

Harold M. CookCharles FendrockHenry G. HeckDavid K. HudsonPaul D. HusbyMark A. LysneRobert W.Mann, Jr.

Richard J. McCarthy

1976

lanis Bestul OssmannJeslie R. ChermakGarret A. DaviesJeffrey J. HeldKelly P Mcddlan

Juzer S. MogriGeorge W. Todd, IVJames P. Wajda

Robert J. Winkler

1977

Paul I. AcknunCharles B. BaltimoreDanny Bieio W.

Earl H. BunkerWalter H. GoodwinBrian G. R. HughesCharles G. Mogged, Jr.Timothy F. MoraonMichael W. SonnenfeldtG. Scott Thompson

1978

Peter C. CoffeeSusan L. KaytonTapio L. KuusinenBarbara K. Ostrom

1979

Anne M. Michon

1980

Timothy M. Folsler

Peter J. Francis

1981Stuart L. AndersonThomas P. GariganWilliam I. OgtlvTe

1982

Robert D. Powell

Aeronautics and Aatronautii

Carl Alexoff '56Daniel H. Daley '46Carlo N. De Genruro '53David W. Dove 71

lames W. Harrill '64Richard D. Linndl '48John A. Long 33Howard A. Magrath '38lames S. McDonnell, III '59

James S. Miller '61Theodore H. H. Pun '48John C. Ryan '60George S. Sduirer "35Leroy P. Smith '49A. Tobey Yu '46

Architecture

Frederick R. Bentel '50BUI C. Boozions '60Thomas D. Cabot. Ill TOC. Rosalie C. Carson '47

Robert P. Cooke '62Yusine Y. S. Jung '62Toufic E. Kadri 112John-W. Peirce '47Ewart A. Wetherill '58

David Baltimore '61Russell Kuo-Fu Chan 74

Barry J. Fry 70

David A. Cubbins '81Peter N. Rosenthal '68Robert H. Rubman 71Trent S. Russell '42Alfred M. Webb '47Barbara A. Wolf 73

Chemical Engineering

Benson U. C. Aghazu 72Leonard Berkowitz '58P. L. Thibaul Brian '56James S. Bruce '39Bernard Chertow '48Jerry A. Cogan 32John P. Cogan 32Robert H. Cotton 39Andre C. DePrez '55John E. Fay, II 71Joel H. Friedman '60Maurice F. Granville 39Robert L. Greene '47Hugh Robert James 74George R. Jasny '52Earp F. Jennings, Jr. 39

James R. Katzw 70Ernest 1. Korchak '61James Lago '47Edward A. Mason '50Jerry McAfee '40John E. Millard 35

Marion Monet '43Tim Montgomery 74Edward W. S. Nicholson 36

John H. O'Neill, Jr. '51R. Robert Paxton '49

Donald W. Peaceman '51

William A. Reed '43Robert L. Richards. Jr. '51Keith E. Rumbd 70Leonard W. Russum '47George F. Schlaudecker 39John P. Schmidt'63

Hugh W. Schwarz '42Ronald A. Shulman 'S7Robert E Siegfried '47Robert S. Smith '47Herbert L. Stone '53William E. Tucker, Jr. '42Cheng C. Wang '46Douglass I. Warner '59

James C. Wei 'S4

Martin A. Welt 'S7

Jack C Williams 38

KwangJ. Won 79Irwin 5. Zonis '52

ChemistryFred A. Bickford '33James J. Bishop '69Clifton J. Blankley '67Cart H. Brubaker. Jr. '52Howard S. Carey, Jr. '55Hugh L. Dryden. Jr. '50Lionel S. Galstaun '34Harbo Peter Jensen 74Andrew V. Nowak 72la net Sanford Perkins '52

Emily L. Wick '51

Civil EngineeringThomas W Anderson 37

Albert H. Bryan. Jr. '48Shelby H. Curlec '60Kenneth C. Deemer '52David D. Draco!I '69

M. David Egan '66Thomas F. Gilbane. Jr. 75Charles W. Johnson '55Thomas D. Landate '54Shih Y. Lee '43Frederick G. Lehman 39William O. Lynch '47Thomas S. Maddock '51William A. Moytan '80John F. CTLeary '66Kwan D. Park *S5

i Arthur C. Ruge 33Hie A. Sehnaoui '61

William C Stookey '51

Richard A. Sullivan '59George P. Turn '56Charles R. Walker '48Stanley M. White 76Roger H. Wingate 37

Earth and Planetary SciencesNicholas G. Dumbros 34Louise H. Herrington '41

Economio

Robert W. Adams '51Leslie Cookenboo 'S3Vincent A. Fulmer '53Thomas G. Hall'52

J. Wade Miller '48

Electrical EngineeringArthur H. Ballard'50T

David R. Bieberle 78Richard W. Boberg 73Anthony P. Di Vincenzo "47jay W. Forrester '45Hans P. Geering 71Robert F. Hossley 73Stephen J. latras '52

Alexander Kusko '44Jean D. LebeJ S3Gordon M. Lee '44Henry S. Magnuski 73Robert L. Massard '50

Terrence P. McGarty, Jr. 71Charles W. Merriam, III '55John Richard Mulhern 70Cedric F. ODonndl '51Robert A. Price 'S3Alexander L. Pugh. Ill 'S3James R. Rdyea*58William M. Snyder. Jr. 39

John R. Whitford '49

ManagementGeorge A. Bobelis '58Roy O. Brady, Jr. 72Ralph N. Bussard '69Vincent S. Castdlano 77

Robert V. dapp '63William L. afifon, Jr. 70John F. Fort, 111 '66

Donald V. Fowke '63Maurice A. Garr, Jr. '49Kenneth F. Gordon '60Donald M. Hague 77

Frederick L. Hill '67Winston R. Hindle. Jr. '54Jack Hubbard '63Thomas G. loerger 75Stephen A. Landon '68James B. Law '52Paul H. Levy 77

John Norris Maguire '60Robert Y. Mao 72Karen Mathiasen 71

Bruce A. H. McFadden 75Robert Barry Moser 74Donald H. Peters '69John D. Proctor 73

Arnold O. Putnam '47Thomas J. Quinlan HIRichard G. Schweikhardt 73

Arthur!. Seiler 32Lee L. Selwyn '69Denis M. Slavich 71William E. Small '69Dan S. Somekh '67

James A. F. Stone! '61Allen N. Strand '64Susan K. Sudman 75

Yaron D. Teplow '62Martin Trust '58

Thomas R. Williams '54Zenon S. Zannetcs '55

Materials Scienceand Engineering

J. Howard Beck 35

F. William Bloecher '49

John A. Fellows 32George E. Nereo '63Robert C. Ruhl '67Reinhardt Schuhmann, Jr. 38Howard R. Spendelow. Jr. '42

JanineJ. Weins 70

Mathematics

Alan E. Berger 72David A. Castanon 76Robert A. dark '49

John H. Doles. Ill 69Andrew M. Odlyzko 75Janice R. Rossbach '51Robert E. Sacks 75Claude E. Shannon '40Norton Starr '64

Mechanical Engineering

Anthony E Alonzo '58John C. Chato '60Robert H. Davis '50Charles N. Griffith! 37

Hdge K. Heen '55G«nge E Keeler '54George A. Lavoie 70Roger L. McCarthy 77Ma°len L. Miller '54Watson E. Siabaugh 30Jan A. Vdtrop 'S3Chiao J. Wang '46

MeteorologyJames R. Mahoney '66Leonard W. Weis '47Naval Construction

and EngineeringNorman K. Berge '60Charles A. Curtze 38Dean A. Horn '49Frank Cox Jones '43David R. Saveker '46

Nuclear Engineering

Harry J. Capossda 38William R. Corcoran 71Dale E. Crane '67Michad J. DrisccJI '64 'Victor I. Orphan '64Philip F. Palmedo '58Richard E. Price '63Tadeusz J. Swierzawski '62Neil E. Todreas '66Ian B. WaU '64

Nutrition and Food SdenceCharles J. Bates '57Darshan S. Bhatia 'SOGeorge L. Blackburn 73Jean-Louis Fribourg 72Lawrence D. Starr 55David H. Wallace '63

Ocean EngineeringHugh H. Fuller, iff 73Edwin MaUoy, Jr. '45

Physio

Solomon J. Buchsbaum '57Eugene I. Gordon '57F. 5. Holmes, Jr. 73Lincoln B. Hubbard '67Norman C. Rasmussen '56

David J. Rose'»Parr A. Tate 'S3Glenn R. Young 77

Senior Executive* Program

Mike FUlipoff 31Andrew C. Knowles 76lames D. Reeves. Jr. 71Richard L. Terrell 58

Yoshito S. Yamaguchi "82

Sloan Fellows Program

Rudolph Alexander "81Robert S. Ames '54

Robert B. Anderson '66Ray W. Ballmer '60

Eugene D. Becken '52Louis P. Bodmer '57Robert E. Bryan, Jr. '69Richard A. Burke 79Wayne H. Burl '58Daniel F. Cameron '59Albert T. Camp '56Robert H. Campbell 78

Steve Cenko '64Franco Qiiesa TCWendel W. Cook '68John C. Davis '56Richard A. DeCcate 77William R. De Long '60Donald A. Dick '6ffPeter D. Fenner 75Carmen Ferranli 70Henry E Fish '61Reinhard Frank 74Stuart M. Prey '61Leonard W. Golden '55

Charles R. Grader 74Walter K. Graham 39

Frank H. Hall. Jr. '42Donald A. Henhksen '68Takashi Iwamura 78

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Ronald W. O'Connor 71

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UndetignatedPhilip A. Le Bar, Jr. '69RlchardS. McCurdy 70Elizabeth J. YeaIrs 74

Urban Studies and PlanningTheodore S. Bacon. Jr. '56Samuel M. Ellsworth '55Allen G. Gerstenberger 74

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HonoraryRoy LamsonJames N. Phinney

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Rocliord. IL61I07

TECHNOLOGY REVIEW A21

Solutions

M/J 1. White to move and mate in four.

Elliot Roberts sent us the following solution:

... K-r7

2. ... if KxN

3. R-d7 . . . K-h5

4. R-h8 mate

2. N-h6 ... if K-h7

3. K-f6 . . . KxN

4. R-h8 mate

Also solved by Edward Gaillard, Richard Hess,

Peter Hagelstein, Everett Leroy, Ronald Raines,

David Evans, Matthew Fountain, and the proposer,

John Cronin.

M/J 2. Find a nontrivial, continuous, real-valued

function f(x) possessing continuous derivatives of

all orders and satisfying the infinite-order differen

tial equation:

f = V + 4f" + 9f" + 16f"" + . . .

Paul Schweitzer writes:

A solution is

f(x) = Ae" (1)

where A is real and arbitrary, and where a is a real

zero of the polynomial g(x) =1 — 4x + 2x* — xa,

lying between —1 and 1. To see this, substitute (1)

into the differential equation

0 = -f + Df + 4D»f + 9DJf + 160*1 + . . . (2)

to get

- y n'a0 = Ae*

(sum convergent for |a| < 1)

= Ae'

Ae"

a(l + a) 1(1 -a)»J

Consistent if g(a) = 0. Since g(.25) = 7/64 > 0 and

g(.3) = -.047 < 0, a lies between .25 and .3. Also,

since dg(x)/dx = 4(x - 1) - 3x' is ^ 0 for x s 1, this

is the only real choice of a: inside 1 -1, 1) with g(a)

= 0.

Also solved by Henry Lieberman, Leo Harten,

Charles Sutton, John Prussing, Jerry Grossman,

Stephen Persek, Richard Askey, Matthew Foun

tain, Peter Hagelstein, Richard Hess, and the pro

poser, Frank Rubin.

M/J 3. Two cryptarithmetic puzzles:FVFR

= .ONANDONANDON. .NNNNN

EVER

VVVW• .ONANDONANDON.

IRIS

Hill

Robert Way found 767803333 as a solution to the

first part and proved the second has no solution.

The proposer offers 3638/66666 but does not com-

Ken

Eldred

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TECHNOLOGY REVIEW A23

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m.2

mb2

m,2

= h,2

= hV

= h,2

1

U

y)'

•r

X1

y'

z2

II

=b2-

= c1 -

= a2-

h6"

h>2

K'

(a

(b

(c

-x)2

-y)'

-z)'

III

= c' -h.2

= a2-h.2

= b2 - h€2

ment on IRIS/IIIH. Perhaps I have misinterpreted

the problem.

Also solved by Richard Hess, Naomi Markovitz,

and Matthew Fountain.

MJJ 4. Find three consecutive three-digit prime

numbers x, y, z, such that z - y = y - x = 12. In

other words, there are no prime numbers between

x and z, other than y. Find four consecutivethree-digit prime numbers a, b, c, d, such that d -

c = c-b = b-a = 6. What is the largest string of

consecutive primes P,, P., . . ., Pn such that:

l\ - P, , = Pn-, - Pn-j = . . . = P, - P, = 4.Charles Sutton writes:

I found this problem particularly interesting, since

the basic idea involved, consecutive primes in

arithmetic progression, can be considered a

generalization of the well-known twin prime prob

lem. One would expect the distribution of pairs ofprimes differing by 4 to be similar to the distribu

tion of the twin pairs, of which there is conjectured

to be an infinite number. Are there also infinitely

many sets of three or more consecutive primes in

arithmetic progression with a given common difference? An intriguing question!

We are asked to find a set of three consecutivethree-digit prime numbers in arithmetic progres

sion with common difference 12. One such set is

found to be 199, 211, 223. Again, we arc asked to

find a set of four consecutive three-digit prime

numbers in arithmetic progression with common

difference 6, and the set 251,257, 263, 269 is found.The last question asked is what is the longest

string of consecutive prime numbers in arithmetic

progression with common difference 4. The sur

prise answer is that the longest possible string con

sists of only two primes. Consider the first member

of such a string. Since it is prime, its remainder

when divided by 3 must be either 1 or 2. If the re

mainder is 1, then the next prime in the string (just

add 4 to (he first) will leave a remainder of 2 when

divided by 3, but the third will be exactly divisibleby 3, so it can't be prime. Thus the length of the

string is two. And if the remainder when the first

member of the string is divided by 3 is 2, then the

string will have length one, since the next number

(add 4) will be divisible by 3, so it can't be prime.

Also solved by Phelps Meaker, Frank Carbin, R.

Way, Sidney Feldman, David Evans, Peter Hagels-

tcin, Leo Hartcn, Richard Hess, Matthew Fountain,and the proposer, Larry Bell.

Ml) 5. For any triangle, the sum of the squares of

the medians to the sum of the squares of the sidesequals a constant rational number. What is thisnumber?

Here is what Farrel Powsner describes as "the

most publishablc solution":

mc to sides a, b, and c, respectively. (In order to

simplify the diagram, m. and ha are the only segments interior to the triangle that are drawn.)

mi, and ho separate C_A into segments whose

lengths are

b b- . j - y, and y.

mc and ht separate AT into segments whoselengths are

— , — — z, and z.

Using the Pythagorean Theorem gives the resultsshown in the box above.

Summing II and III and setting the results equal:x2 + y2 + z2 = (a - x)2 + (b - y)2 + (c - z)2x2 + y2 + z2 = (a2 + b2 + c2) + (x1 + y2 + z") -2(ax + by + cz)

2(ax + by + cz) = a2 + b1 + c2

ax + by - a = Vi(a2 + b2 + c2). (1)

Summing I and expanding:

m.2 + itu2 + mt! = (ha2 + h,,2 + h02) +

a2 b2 '■^a2 b2

(ax + by + cz).

Now, summing U:

x2 + y2 + z2 = (a2 + b2 + c') -

(h.2 + h,,2 + hc2).

Substituting (1) and (3) into (2):

m,2 + nu2 + mc2 = a1 + b2 + c2 +

V«(a2 + b2 + cJ) - Vj(a2 + b2 + c2).

Therefore,

ma2 + mi,2

2 2

(2)

(3)

+ mc! = + b2 + cl). And

a2 + b2 + c2

Therefore, the ratio of the sum of the squares of the

medians of any triangle to the sum of the squares ofthe sides is %.

Also solved by Mary Lindenberg, Jack Hiatt, John

Prussing, Raymond Cailbrd, Norman Spencer,

Ken Haruta, Steve Shapiro, Frederick Furland, R.

Way, Naomi Markovitz, Phelps Meaker, SidneyFeldman, Peter Hagelstein, David Evans, Henry

Lieberman, Ronald Raines, Richard Hess, Matthew

Fountain, and the proposer, Harry Zarcmba.

Belter Late Than Never

F/M 4, F/M 5 Raymond Caillard has responded.

Proposers' Solution to Speed Problems

OCT SD1. All equal their logs except for position

of decimal point [I didn't get it either— Ed.).

OCTSDZ

4- B

In the diagram, we have triangle ABC with

altitudes h., h,,, and h, and medians m,. m,,, and

24-13

13-5

13-11

J-13

24-13

13-5

5-24

A24 OCTOBER 198J

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Edible Checkers and

a Freshman's Folly

A lovely letter from Mary Lindenberg

reports that the M.I.T. and Hunter Col

lege alumni associations each met lastJune 10 at M.I.T.—and she was part ofboth events. Ms. Lindenberg has anunusual double career; she is a teacherof mathematics and also of painting.She has recently sent us a reprint of her

article on originality and creativity—topics germane to both disciplines.

Problems

N/D 1 We begin with a chess problemfrom George Farnell:

White is to play and mate in three.

N/D 2 Mitchell Serota recalls that cancelling d's to give

dy/dx = y/x

is known as "freshman's folly." Butsimilar jokes occasionally work forarithmetic:

64/16 = 4/1

326/163 = 2/1

Can anyone find a four-digit counterpart?

N/D3 Niles Ritter and Andrew Bernoff

first sent this problem to M3, the M.I.T.math majors' magazine (why not call itM»?)

At the Mathematics Department tea, Ihad a collection of equal numbers of redand black checkers, and when the con

versation grew dull 1 discovered that my

Puzzle Comer/Allan J. Gottlieb

Allan). Gottlieb, '67, isassociate researchprofessor at the CourantInstitute of

Mathematical Sciencesof New York University;he studied mathematicsat M.I.T. and Brandeis.Send problems,solutions, and

comments to him at the

Courant Institute, NewYork University, 251Mercer St.. New York.N.Y. 10012.

checkers could be arranged in a rectangular array, with all the black ones on

the border, like this:

O O O ••••

o o o

o o o

o

o

o

o o

o o

o o

o o o

o o o

o o o

o o o • •• •

o o o • •• •

o o o • • • •

But when my back was turned, one of

the professors ate a few of each color,thinking my checkers to be cookies. I

found that 1 still had an equal number ofblack and red checkers, and that 1 couldstill arrange them in an array with thereds inside and the blacks on the border. How many checkers did I startwith, and how many were eaten?

N/D 4 This problem, described as "a

curiosity from the field of speed indica

tion," first appeared in Technology Re

view in 1938 as part of an advertismentfor Calibron Products, Inc."

A hemispherical bowl with a radius of 1foot 1 inch is mounted on a central vertical shaft YY. A one-pound ball B with a

radius of 1 inch is free to roll inside thebowl. On what part of the bowl's surface will the ball tend to ride (i.e., what

will be the value of x) if the bowl is spun

at 50 R.P.M. about the YY axis? How do

A30 NOVEMBER/DECEMBER 1983

taught physics at Phillips Exeter

Academy, and later he was a member of

the faculty at Wayne University. His

work with PSSC led to a number of

books by members of the project and to

several films, including "An Introduc

tion to Optics" which is still used as a

teaching aid.

Deceased

John Hall, '04; July 13, 1983; 3838 Halifax Rd..

Wilmington, N.C

Maurice H. Pease, '07; September 10, 1983; 5093

Starfish Ave., Naples, Fla.

John S. Barnes, '08; 1980; 18 Woodland Rd.. Law-

rencevillc, N.J.Mrs. Earl W. Pilling, '10; 1983; 767 Washington St..

Norwood, Mass.

Uoyd C. Cooley, '11; October 2. 1982; Plymouth

Harbor Apt. 2205. 700 John Ringling Blvd.,

Sarasota, Fla.

Clarence R. Woodward, '12; February 24,1983; 905

Country Club Dr., Greensburg, Penn.

Louis C. Rosenberg, '13; June 9. 1983; c/o Mt. View

Convalescent Circle, 1400 Division St., Oregon

City, Ore.

Raymond A. Meader, '17; February 8, 1980; 16522

Burr Hill, do D F Malder, San Antonio, Tex.

James M. Todd, '18; March 23, 1983; 1489 Cbir-

mont PI., Nashville, Tenn.

Dean K. Webster, Jr., 19; July 29, 1983; 27 Royal

Crest Dr., Lawrence, Mass.

Mendum B. Littlcfieid, '20; July 4, 1983; Littleton

House, Littleton, Mass.

Weston Hadden, '21; June 14, 1983; 22 Monument

Ave., Bennington, Vt.

Harry M. Ramsay, '21; May 12, 1983; 11 E Orange

Grove Rd. No. 116, Tucson, Ariz.

Edwin L. Rose, '21; July 7, 1983; PO Box 116, Sierra

Madre, Calif.Haywood P. Cavlarly, Jr., '22; June 25, 1983; 1615

North Oleander Ave., Datona Beach, Fla.

William J. Edmonds, '22; July 1, 1983; 2701 Gulf

Shore Blvd. N., Naples, Fla.

Morris H. Gens, "22; April 29, 1983; 75 Lee St.,

Brookline, Mass.

Charles T. McCrady, '22; May 15, 1983.

John J. Breen, '23; June 6, 1974; 174 Summit Ave.,

Summit, N.J.

Mrs. Kenneth G. Crompton, '23; 1983; 407 Prospect

St., Lawrence, Mass.

Kenneth C. Kingsley, "23; June 9, 1983; 649 Via

Lido Soud, Newport Beach, Calif.Theodore M. Burkholder, 74; August 4, 1983; 60

Summit St., Newton, Mass.

Charles E. Herrstrom, '24; September 1, 1983; 700

John Ringling Blvd. No. 905, Sarasota, Fla.

Jacob A. Manian, '24; April 10, 1983; 39 Acorn Cir

cle Apt. 202, Towson, Md.Samuel Glaser, '25; August 7, 1983; 381 Dudley

Rd., New Centre, Mass.

Samuel J. Cole, '26; May 10. 1983; 11 Wilde Ave.

Apt. 2, Drexel Hill, Penn.

George J. Leness, '26; August 17, 1983; 31 E 79lh

St., New York, NY.

Lucas E. Bannon, '27; March 29, 1983; 19 No. Col

umbus St., Beverly Hill, Fla.

Francis T. Cahill, '27; June 29, 1983; McKoy Rd.,

North Eastham, Mass.

Richard Cutts, Jr., '27; March 3, 1983; 305 Green

wich Ave. Apt. 130B, Warwick. R.I.

Frank G. Kear, '27; July 22. 1983; 501 Portola Rd.

No. 8085, Menlo Park, CalifMrs. Thorwald Larson, '28; October 2, 1982; 1535

Pine Valley Blvd. No. 110, Ann Arbor, Mich.

Robert S. Woodbury, 28; September 18, 1983; 12

Meadowbrook Rd., Dover, Mass.

Russell B. Wright, '28; October 14, 1982; c/o Mrs.

Henry Giugni, 1518 Adams St., Saint Helena, Calif.

J. Gordon Carr, '29; August 10, 1983; 46 Beechcroft

Rd., Greenwich, Conn.

Daniel J. O'Connell, '29; July 14, 1983; 50 Elliot St.,

Holyoke, Mass.

H. Dayton Wilde, '29; July 20, 1983; 3013 Avalon

PI., Houston. Tex.

denis R. Agar, '30; August 2, 1983; 2625 Regina St.No. 1706, Ottawa, Ont., Canada.

William Harold Bethel, '30; August 28. 1982;

Hallmark Nursing Home, 49 Marvin Ave . Troy,

N.Y.

Edward H. douser, '31; September 1956; 308 Druid

Rd., Clearwater, Fla.

Eliot S. Graham, '31; August 1, 1983; 2331A Av-

enida Sevilla, Laguna Hills, Calif.

Michael Kundralh, '31; August 6. 1983; 144 Red

Oak Rd., Fairfickl. Conn.

Edwyn A. Eddy, '32. September 11, 1983; RFD 1,

Winsted, Conn.

John W. Leslie, 32; June 29, 1983; 42 Whitney St.,

Medford, Mass.

Dominic A. Perry, '32; July 7, 1983; 533 Paddock

Ave., Meriden, Conn.

Joseph L. Richmond, 32; 1972.

Robert M. Trimble, '33; March 1983.

George R. Forsburg, '35; August 7, 1983. 78 Clisbv

Ave., Dedham. Mass

John A. Kleinhans, 36; September 13, 1983; 3064

Kent Rd. Apt. 411A, Cuyahoga Falls. Ohio.

Leo F. McKenney, '36; August 19, 1983; Dogford

Rd., Etna, N.H.

Morril B. Spaulding, Jr., '36; 1980; 4341

Montgomery Ave., Belhesda, Md.

Harrison S. Woodman, '36; June 28, 1983; 53 Buena

Vista Ave., Rumson, N.J.

Willard R. Beye, '38, September 3, 1983; 4414 Mar

seilles St., San Diego, Calif.

Francis S. Stein, '38; June 17, 1983; 5 Brooke Ave.,

Annapolis, Md.

Don Cornish, '39; 1981; 17846 Ballinger Way NE,

Seattle, Wash.

Akim S. Zaburunov, '39; June 10, 1983; I'O Box

1703. Fort Collins, Col.

Oliver K. Smith, '40; August 11, 1983; 1025 Us

Pulgas Rd.. Pacific Palisades, Calif.

Merlen C. Bullock, '42; September 17, 1983; 2007

North Medina Line Rd., Akron, Ohio.

Louis V. Sutton, Jr., '42; July 15, 1983; PO Box 3085,

Rock Hill, S.C.

Robert A. Bamford, '43; July 2, 1983; 20 Rene Rd.,

Brockton, Mass.

Frank W. Bailey, '46; 1982; 60 Parkway Dr. E., East

Orange, N.J.

Stuart D. Grandficld, '46; August 23, 1983; 328 Bar

ranca Ave. No. 2, Santa Barbara, Calif.

William W. Caudill, '47; June 25, 1983; 10923 Kir-

wick, Houston, Tex.

Boynton H. Tucker, '49; March 1983; RT 25H Box

34A, Malakoff, Tex.

Arthur H. Schein, '51; September 14, 1983; 22 Puri

tan Rd. Newton Highlands, Mass.

Walter J. K. Tannenberg, '52; July 12, 1983; 1

Longfellow PI., Boston, Mass.

Barton Roessler, '55; June 16, 1983; 4 Indigo Rd.,

Barrington, R.I.

Philip T. Andrews, '57; March 25,1983; 20 Gunder-

son Rd., Wilmington, Mass.

William H. Coghill, '57; December 24, 1982; 1111

5lh Ave., Sebring, Fla.

Miguel A. Barasorda, '59; December 30, 1981; Or-

quidea 12, Sta Maria, Rio Piedras, Puerto Rico.

Sylvia C. Bluhm, '59; December 16, 1981; 1253

Cambridge Ave., Morgantown, W.V.

Robert R. Thompson, '59; November 1978.

C. Morgan Harris, '62; August 1977.

Robert K. Bofah, 71; 1983; I'O Box 10, Goaso,

Ghana.

Harry Boothman, 73; 1977; City of Calgary, Parks

and Recreation Dept., Calgary, Alt., Canada.

Robert U. Sautter, 74; May'7, 1982; 54736 Mer-rifield Dr., Mishauaka, lnd.

Christopher Lawlor, 75; August 25, 1982; 9285

Root Rd., North Ridgeville, Ohio.

William G. McCabe, '77; May 16, 1983; 20 Hillside

Ave., Haverstraw, N.Y.

Stephen D. Holland, 79; September II, 1983;

Baker Bridge Rd., Lincoln, Mass.

Stephen M. PaneiU, 80; September 1. 1983; 8

Goodman Rd., Cambridge, Mass.

Sudhir K. Sarin, '83; August 7, 1983; 80 Ranleigh

Ave., Toronto, On!., Canada.

H.H.Hawkins

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TECHNOLOGY REVIEW A29

you explain this peculiar result? Show

that x will increase to about 7 inches at a

speed of 60R.P.M.

N/D 5 John Rule wants you to find two

five-digit perfect squares that together

contain all ten digits. How many so

lutions exist?

Speed Department

N/D SD 1 A bridge quickie from Doug

Van Patter:

You (playing East) are defending

against a six-heart contract in a high-

stakes game:

North (dummy):

A 9 6 2

V K 4 3

♦ A 8'

* A 9 7 6 2

East:

A K Q J 8 3

♦ J♦ K 10 5

♦ K 8 4 3

Your partner leads the »fr5, taken by the

«fcA. Declarer pulls three rounds of

trump, then leads the 4>Q to your «frK.

Your partner shows out. Can you find a

way to save a lot of money?

N/D SD 2 David Evans has drawn a

diagonal on each of two faces of a cube

so that the diagonals share a vertex.

How large is the angle between the

diagonals?

Solutions

F/M 1 While to male in three:

)UL 1 South is to make six spades, with the opening

lead from West VQ.

A J 9 8 7 2

V 5 4

♦ A 5 4

+ A K 2

A 6 5 4

V Q J 10 9 8 3 2

♦ 9

There are several solutions to this problem,

which was first published in February/March 1983

and corrected in July 1983. Elliot Robert found the

following:

1. Rd6 KxP

2. K f 3 K f S

3. Qf4

1. KfS

2. Kf3 KxP

3. Qf4

Also solved by E. Stout, David Evans, Robert

Way, Matthew Fountain, Charles Rivers, Randy

Kimble, and the proposers. Bob Kimble and Jacques

Labelle.

♦ —

¥76

♦QJ10 876

*96 *AKQ 103*QJ I087V A K

♦ K 3 2

* 5 4

We received a fine analysis from Robert Way:

The experienced bridge player will be disappointed

that the spade trumps are divided five-and-five in

the declarer's hand and dummy, providing no extra

ruffing values. On the other hand, he will note that

an "automatic" squeeze is available in dubs and

diamonds againsl East once West has been stripped

of these suits and a trick has been conceded to

"rectify the count." The latter can be accomplished

by losing a trump trick to West after stripping clubs

and diamonds. The trick comes back via a "ruff and

discard" return and sets up the ultimate "automatic

squeeze" against East. In more mundane terms, the

simplest play follows. Although the opening lead is

specified as the VQ, the hand may be made

against any opening lead, which declarer wins by

following suit with the VA, 4>A, +K, or 4A.

South must first win these tricks plus the 4kK, VK,

and «fcK. For these first seven tricks, both the order

of play and the cards from the other three hands are

not critical, as long as they follow suit. On the

eighth lead. South plays the <fc3, following with the

A2 from North, forcing West to win with a higher

spade. West, having been stripped of all other

suits, must return a heart which North trumps, and

South discards a diamond so as to retain a dub as

the potential squeeze card. (Note that West, having

to return a heart on the ninth trick, has promoted a

spade winner in the North hand which otherwise

would have fallen under a spade winner in the

South hand due to the five-five even distribution,

thus regaining the trick lost at trick eight and also

delaying the tempo of the game so that East is sub

sequently squeezed.) North now leads his last

spade, which is taken by South's AQ; followed by

the VQ on which North discards a dub. The posi

tion after 11 tricks have been played: West holds no

diamonds of dubs. North holds the +A and a

small diamond, South holds a small diamond and a

small dub (!), and East can have only two cards,

too. If both are diamonds. South plays his good

dub and to North's +A; if East retains a dub, he

can hold only one diamond, and after the play of

North's +A the small diamond is good. Note the

problem if the Declarer plays the ♦A (instead of

the ♦K) among the first seven tricks: the squeeze

position is then transferred to North who must win

the last trump trick. This requires Declarer to save

the AJ in the dummy. On the forced heart return at

the ninth trick North discards a diamond and South

trumps with the AQ in order to "unblock" the

spade suit. The A10 is then led to the North's AJand the last spade cashed, discarding a dub from

South. After 11 tricks. North holds a small diamond

and small dub, while South retains the +K and a

small diamond. Likewise, East, reduced to two

cards, cannot protect both diamonds and dubs.

This line of play is forced if the North-South dubs

are interchanged and the East-West clubs areswapped for diamonds. Then declarer must play

both the +A and ♦K plus a high dub to strip

West, requiring the final squeeze to be initiated by

North.

Also solved by Garabed Zaratarian, Robert Lax,

Matthew Fountain, Kenneth Bernstein, Peter De

Florez, Conrad Carlson, Doug Van Patter, Tom

Harriman, Ben Feinswog, Eric Liban, Charles Riv

ers, and the proposer, Winslow Hartford.

JUL 2 The number of ways the play of a hand of

bridge can proceed obviously depends on the distribution and the choice of trump. Can some rea

sonably tight upper or lower bounds be found?

How about an average case analysis?

Matthew Fountain points out an interesting am

biguity. Consider a hand in which each hand con

sists of 13 cards of the same suit. For any contract,

each player may legally play his 13 cards in any

order. Thus we have 13!' possibilities, dearly an

upper bound. However, in some sense all these

ways of playing the hand are equivalent, so

perhaps this deal should be considered an obvious

lower bound with only one possibility. However,

since this is a combinatorial and not a bridge prob

lem, we will not try to define equivalent hands.

Thus we have an upper bound.

Winslow Hartford has an idea for a lower bound.

The lower limit is for a 4-3-3-3 hand, all around, in

which one player holds say AAKQJ VAKQ

♦AKQ AAKQ. This will give the fewest choices of

play as the winning hand is randomly pbyed out.

Thus there are only 432 ways of playing the first

trick, as opposed to 28,561 for the case above. This

is instictive, but it should be about correct.

Also solved by Robert Way, Tom Harriman, and

the proposer, Jerry Grossman.

JUL 3 Two cryptarithmetic puzzles: Nine is a

square, and while NINETEEN isn't actually a

prime, at least its smallest factor is greater than 150.

TWO + TWENTY = TWELVE + TEN. (The first

three are all divisible by their namesakes.)

The following solution is from Charles Rivers:

NINE is a square and NINETEEN has no prime

factor smaller than 150. Therefore N is dearly an

odd number. Of the four-digit squares (32* through

991), only four have equal first and third digits, and

of these, only one (561 = 3136) has equal first and

third odd digits. Therefore:

NINE = 3136

NINETEEN = 3136X663, where X = 0,2,4,5,7,8,9

Substitution of each possibility in turn and factor

ing gives the following results:

X = 02 45 78 9

Smallest factor of NINETEEN = 89 3 11 3 617 3 13

Therefore, NINETEEN = 31367663 with factors of

617, 50839.

The second part of the problem is somewhat more

difficult:

1. Since TWO is divisible by 2, O is an even

number.

2. Since TWENTY is divisible by 20, Y=0and T is an

even number.

3. Since TWELVE i$ divisible by 12 (and all its fac

tors induding 4), E is an even number with the

further restriction that if E-0,4,8, V is even, and if

E=2,6, V is odd.

4. O + Y = O, an even number with no carryover.

Therefore E + N equals an even number, and since

E is even, N is also even.

5. All even numbers are accounted for

(Y=0,T,O,E,N) so L,V,W are odd and E=2 or 6 (see

3 above).

6. From 4 above, T+W is odd (no carryover). There

fore E + N must be less than 10 (no carryover) so

that E + V + carryover (if any) is odd. Therefore: E

+ N = O and O = 6 or 8; E = 2 or 6:

Y O E N T

0 6 2 4 8

0 8 2 6 4

0 8 6 2 4

7. From all this, TW + NT = LV+TEandL = N +

1.

8. Rearranging and solving for W - V:10T + W + ION + T = 10L + V + 10T + E

W - V = 10(L-N) + E - T

9. Therefore

YOENTL WA; WV TWELVE

062485 473 872532 - 12 = 72711

082647 891 492712 - 12 = 41059.333

086241 -8 1 9 W * L

10. The solution is:

876 872532

872480 824

873356 = 873356

Also solved by Matthew Fountain, David Evans,

Kenneth Bernstein, Fred Furland, Carlyn Iuzzolino,

Winslow Hartford, Robert Way, Sidney Feldman,

John Spalding, W. Woods, Harry Zaremba, Frank

Davis, John Rule, and the proposer, Avi Ornstein.

TECHNOLOGY REVIEW A31

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When Anthony Stanton asked "Puzzle

Comer" readers (JUL 4) about the

geometry of a slotted block with

travelers in the slots, Martin Buerger,

emeritus professor of mineralogy and

crystallography at M.I.T., realized that

JUL 4 A block of wood has slots AB and CD, as

shown. In each slot is a wood piece, P and Q, that

can slide in it. Also shown is a bar attached by loose

screws to P and Q and extended to a handle H. Is it

possible to rotate the handle through a full 360°

without having P and Q collide? tf this is possible,

is the orbit of H an elipse?

The blocks cannot collide since they must remain

a distance PQ apart. Kelly Woods shows us that the

orbit of H is indeed an ellipse:

Let the distance PQ be designated by L and the

distance QH by M. Then, as shown in the diagram.

Stanton's device was essentially

equivalent to this machine—Buerger's

invention—for drawing ellipses.

Buerger's description of the machine is

available on request from the editors.

y = (M + L) sin 6

x = M cos 6

The equation of an ellipse is:

(x/a)J + (y*)> = 1

In this case the minor semi-axis is a = M and the

major semi-axis is b = (M + L), from which:

cos1 9 + sin2 9 = 1.

Martin Buerger, professor emeritus at M.I.T.,

constructed such a device in the 1940s, a photo

graph of which appears above. Professor Buerger

has supplied a written description of the machine,

which can be obtained from the editor upon re

quest. And Norman Wickstrand notes that Keuffel

and Esser once sold such an instrument.

Also solved by Kenneth Bernstein, Matthew

Fountain, Tom Harriman, Fred Furland, Eric Liban,

Winslow Hartford, Waller Moore, Gary Heiligman,

John Prussing, Phelps Meaker, Harry Zaremba,

Frank Davis, and Robert Way.

JUL 5 The following list is the first 19 perfect

squares containing two distinct digits and not con

taining a zero. What are the next two elements of

the sequence:

4x4 = 16 22x22 = 484

5 x 5 = 25 26 x 26 = 676

6x6 = 36 38x38 = 1444

7x7 = 49 88x88 = 7744

8x8=64 109 X 109 = 11881

9 x 9 = 81 173 X 173 = 29929

11 x 11 = 121 212 X 212 = 44944

12 X 12 = 144 235 x 235 = 55225

15 X 15 = 225 264 X 264 = 69696

21 x 21 = 441

Carlyn luzzolino found:

3114* = 9,696,996

81619' = 6,661,661,161

Also solved by Matthew Fountain, David Evans,

Kenneth Bernstein, Robert Way, E. Stout, Winslow

Hartford, Italo Servi, Frank Carbin, Tom Harriman,

and the proposer. Nob Yoshigahara.

Better Late Than Never

M/J2 Garabed Zarbtrian has responded.

M/J5 Arvi Ornstein has responded.

Proposers' Solutions to Speed Problems

N/D SD1 Put the +K on the table. This may lose a

diamond trick, but it guarantees that the Declarer

won't be able to pitch losers on Dummy's last two

dubs. (Declarer's *J is still blocking the suit.)

N/D SD2 60°. A third diagonal can be drawn form

ing an equilateral triangle.

A32 NOVEMBER/DECEMBER 1983